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AN  ELEMENTARY  TREATISE 

ON  THE 

DIFFERENTIAL  AND  INTEGRAL 
CALCULUS 


WITH  NUMEROUS  EXAMPLES. 


BY 

EDWARD  A.  BOWSER,  LL.D., 

Late  Professor  of  Mathematics  and  Engineering 
in  Rutgers  College. 


Total  Issue,  Sixteen  Thousand 


NEW  YORK 

D.  VAN  NOSTRAND  COMPANY,  Inc. 

250  Fourth  Avenue 


Copyright,  1880,  1905, 

BY 

E,  A.  BOWSER 


■0 


Printed  in  the  United  States  oe  America 


Engineering  Library 


SI  7.  2^ 

3 ri  %7!>  E 


PREFACE, 


C-tfvo , 


tHE  present  work  on  tlie  Differential  and  Integral  Calamus  is 


designed  as  a text-book  for  colleges  and  scientific  schools.  The 
aim  has  been  to  exhibit  the  subject  in  as  concise  and  simple  a manner 
as  was  consistent  with  rigor  of  demonstration,  to  make  it  as  attractive 
to  the  beginner  as  the  nature  of  the  Calculus  would  permit,  and  to 
arrange  the  successive  portions  of  the  subject  in  the  order  best  suited 
for  the  student. 

I have  adopted  the  method  of  infinitesimals,  having  learned  from 
experience  that  the  fundamental  principles  of  the  subject  are  made 
more  intelligible  to  beginners  by  the  method  of  infinitesimals  than  by 
that  of  limits , while  Lu  the  practical  applications  of  the  Calculus  the 
investigations  are  carried  on  entirely  by  the  method  of  infinitesimals. 
At  the  same  time,  a thorough  knowledge  of  the  subject  requires  that 
the  student  should  become  acquainted  with  both  methods  ; and  for 
this  reason,  Chapter  III  is  devoted  exclusively  to  the  method  of 
limits.  In  this  chapter,  all  the  fundamental  rules  for  differentiating 
algebraic  and  transcendental  functions  are  obtained  by  the  method  of 
limits,  so  that  the  student  may  compare  the  two  methods.  This  chap- 
ter may  be  omitted  without  interfering  with  the  continuity  of  the 
work,  but  the  omission  of  at  least  the  first  part  of  the  chapter  is  not 
recommended. 

To  familiarize  the  student  with  the  principles  of  the  subject,  and 
to  fix  the  principles  in  his  mind,  a large  number  of  examples  is  given 
atf  the  ends  of  the  chapters.  These  examples  have  been  carefully 
selected  with  the  view  of  illustrating  the  most  important  points  of 
the  subject.  The  greater  part  of  them  will  present  no  serious  diffi- 
culty to  the  student,  while  a few  may  require  some  analytical  skill. 


363459 


iv 


PREFACE. 


In  preparing  this  book,  I have  availed  myself  pretty  freely  of  the 
•writings  of  the  best  American  and  English  and  French  authors. 
Ma  ay  volumes  have  been  consulted  whose  titles  are  not  mentioned, 
as  credit  could  not  be  given  in  every  case,  and  probably  I am  indebted 
to  these  volumes  for  more  than  I am  aware  of.  The  chief  sources 
upon  which  I have  drawn  are  indicated  by  the  references  in  the  body 
of  the  work  and  need  not  be  here  repeated.  For  examples,  I have 
drawn  upon  the  treatises  of  Gregory,  Price,  Todhunter,  Williamson, 
Young,  Hall,  Rice  and  Johnson,  Ray,  and  Olney,  while  quite  a num- 
ber has  been  taken  from  the  works  of  De  Morgan,  Lacroix,  Serret, 
Courtenay,  Loomis,  Church,  Byerly,  Docharty,  Strong,  Smyth,  and 
the  Mathematical  Visitor;  and  I would  hereby  acknowledge  my 
indebtedness  to  all  the  above-named  works,  both  American  and 
foreign,  for  many  valuable  hints,  as  well  as  for  examples.  A few 
examples  have  been  prepared  specially  for  this  work. 

I have  again  to  express  my  thanks  to  Mr.  R.  W.  Prentiss,  Fellow 
in  Mathematics  at  the  Johns  Hopkins  University,  for  reading  the  MS. 
and  for  valuable  suggestions. 

E.  A.  B. 


Rutgers  College, 

New  Brunswick,  N.  J.,  June,  1880 


NOTE  TO  THE  TWENTY-FIRST  EDITION. 

Tnis  edition  is  enlarged  by  640  additional  examples.  The 
author’s  thanks  are  due  Mr.  Irving  S.  Upson  and  Prof.  William  E. 
Breazeale  for  reading  the  proof-sheets,  and  to  the  latter  gentleman 
for  verifying  many  of  the  answers. 

New  Brunswick,  June,  1905. 


TABLE  OF  CONTENTS 


PART  I. 

DIFFERENTIAL  CALCU  LUS, 


CHAPTER  I. 

FIRST  PRINCIPLES. 

&ET.  PAGE 

1.  Constants  and  Variables 1 

2.  Independent  and  Dependent  Variables L 

3.  Functions.  Geometric  Representation 2 

4.  Algebraic  and  Transcendental  Functions 4 

5.  Increasing  and  Decreasing  Functions 5 

6.  Explicit  and  Implicit  Functions 6 

7.  Continuous  Functions 6 

8.  Infinites  and  Infinitesimals 7 

9.  Orders  of  Infinites  and  Infinitesimals 8 

10.  Geometric  Illustration  of  Infinitesimals 10 

11.  Axioms 12 

Examples 13 

CHAPTER  II. 

DIFFERENTIATION  OF  ALGEBRAIC  AND  TRANSCENDENTAL 
FUNCTIONS. 

12.  Increments  and  Differentials 15 

13.  Consecutive  Values 16 

14.  Differentiation  of  Sum  of  a Number  of  Functions 16 

15.  To  Differentiate  y = ax  ±b 17 


O/TO/B  CQ 


VI  CONTENTS. 

abt.  pass 

16.  Differentiation  of  a Product  of  Two  Functions 18 

17.  Differentiation  of  a Product 20 

18.  Differentiation  of  a Fraction 20 

19  Differentiation  of  any  Power 21 

Examples 23 

Illustrative  Examples 26 

20.  Logarithmic  and  Exponential  Functions 29 

21.  Differentiation  of  an  Exponential 31 

22.  Differentiation  of  an  Exponential  with  Variable  Base 32 

Examples 33 

23.  Logarithmic  Differentiation.  Examples 34 

Illustrative  Examples 32 

TRIGONOMETRIC  FUNCTIONS. 

24.  To  Differentiate  y = sin  x 37 

25.  To  Differentiate  y = cos  x 38 

26.  To  Differentiate  y — tan  x 88 

27.  To  Differentiate  y = cot  x 39 

28.  To  Differentiate  y = sec  x 39 

29.  To  Differentiate  y = eosec  x 40 

30.  To  Differentiate  y — versin  x 40 

31.  To  Differentiate  y — covers  x 40 

32.  Geometric  Demonstration 41 

Examples 43 

Illustrative  Examples 44 

CIRCULAR  FUNCTIONS. 

33.  To  Differentiate  y = sin-1  x 46 

34.  To  Differentiate  y = cos-1  x 46 

35.  To  Differentiate  y = tan-1  x 47 

36.  To  Differentiate  y = cot-1  x 47 

37.  To  Differentiate  y — sec-1  x 47 

38.  To  Differentiate  y = cosec-1  x. 47 

39.  To  Differentiate  y = vers-1  x 48 

40.  To  Differentiate  y = covers-1  x 48 

Examples 48 

Miscellaneous  Examples 49 


CONTENTS.  Vil 

CHAPTER  III. 

LIMITS  — DERIVED  FUNCTIONS. 

iB"  FAGB 

41.  Limiting  Values 59 

42.  Algebraic  Illustration 59 

43.  Trigonometric  Illustration 60 

44.  Derivatives 62 

45.  Differential  and  Differential  Coefficient 63 

46.  Algebraic  Sum  of  a Number  of  Functions 63 

47.  Product  of  Two  Functions 64 

48.  Product  of  any  Number  of  Functions 65 

49.  Differentiation  of  a Fraction 66 

50.  Any  Power  of  a Single  Variable  67 

51.  Differentiation  of  log  x 68 

52.  Differentiation  of  ax 68 

53.  Differentiation  of  sin  x 68 

54.  Differentiation  of  cos  x 69 

CHAPTER  IV. 

SUCCESSIVE  DIFFERENTIALS  AND  DERIVATIVES. 

55.  Successive  Differentials.  Examples 71 

56.  Successive  Derivatives 74 

56a.  Geometric  Representation  of  First  Derivative.  Examples  . 76 

CHAPTER  V. 

DEVELOPMENT  OF  FUNCTIONS. 

57.  Definition  of  Development  of  a F unction  81 

58.  Maclaurin’s  Theorem 81 

59  The  Binomial  Theorem 85 

60.  To  Develop  y = sin  x and  y = cos  x 85 

61.  The  Logarithmic  Series 86 

62.  The  Exponential  Series 90 

63.  To  Develop  y = tan-1  x 91 

64.  Failure  of  Maclaurin’s  Theorem . 92 

65.  Taylor’s  Theorem.  Lemma 93 

66.  To  find  Taylor’s  Theorem 95 

67.  The  Binomial  Theorem 97 


Viii  CONTENTS, 


tET  PASS 

68.  To  Develop  u'  = sin  (x  + y) 98 

69.  The  Logarithmic  Series 98 

70.  The  Exponential  Series 98 

71  Failure  of  Taylor’s  Theorem 99 

Examples 100 


CHAPTER  VI. 

EVALUATION  OF  INDETERMINATE  FORMS. 


72.  Indeterminate  Forms 103 

73.  Common  Factors.  Examples 104 

74.  Method  of  the  Differential  Calculus 105 

GO 

75.  To  evaluate  Functions  of  the  form  — 108 

co 

76.  To  evaluate  Functions  of  the  form  Oxco Ill 

77.  To  evaluate  Functions  of  the  form  00  — 00 112 

78.  To  evaluate  Functions  of  the  forms  0°,  00  °,  and  1±CD 113 

79.  Compound  Indeterminate  Forms 116 

Examples 116 


CHAPTER  VII. 


FUNCTIONS  OF  TWO  OR  MORE  VARIABLES.  — CHANGE  OF 
THE  INDEPENDENT  VARIABLE. 

80  Partial  Differentiation 120 

81.  Differentiation  of  a Function  of  Two  Variables 122 

82.  To  find  the  Total  Derivative  of  u with  respect  to  2: 125 

83.  Successive  Partial  Differentiation . 130 

84.  Proof  that  Order  of  Differentiation  is  indifferent 131 

85-  Successive  Differentials  of  a Function  of  Two  Independent 

Variables 133 

86.  Implicit  Functions 135 

87.  Differentiation  of  an  Implicit  Function 136 

88.  Second  Derivative  of  an  Implicit  Function 138 

89.  Change  of  the  Independent  Variable 14C 

90.  General  Values  of  % , etc 141 

dx  dx-  dx? 

91.  Transformation  for  Two  Independent  Variables 145 

Examples 147 


CONTENTS. 


IX 


CHAPTER  VIII. 

MAXIMA  AND  MINIMA  OF  FUNCTIONS  OF  A SINGLE 
VARIABLE. 

AUT.  PASS 

92.  Definition  of  a Maximum  and  a Minimum 151 

93.  Condition  for  a Maximum  or  Minimum 151 

94.  Geometric  Illustration 152 

95.  Discrimination  between  Maxima  and  Minima 154 

96.  Condition  given  by  Taylor’s  Theorem 154 

97.  Method  of  finding  Maxima  and  Minima  Values 155 

98.  Alternation  of  Maxima  and  Minima  Values 156 

99.  Application  of  Axiomatic  Principles 157 

Examples 159 

Geometric  Problems 164 

CHAPTER  IX. 

TANGENTS,  NORMALS,  AND  ASYMPTOTES. 

100.  Equations  of  the  Tangent  and  Normal. 172 

101.  Length  of  Tangent,  Normal,  Subtangent,  etc 175 

102.  Polar  Curves.  Tangents,  Normals,  Subtangents,  etc 178 

103.  Rectilinear  Asymptotes  181 

04.  Asymptotes  determined  by  Expansion 184 

105.  Asymptotes  in  Polar  Co-ordinates.  Examples 186 

CHAPTER  X. 

DIRECTION  OF  CURVATURE — SINGULAR  POINTS — TRACING 
OF  CURVES. 

106.  Concavity  and  Convexity 191 

107.  Polar  Co-ordinates 192 

108.  Singular  Points 194 

109.  Points  of  Inflexion  194 

110.  Multiple  Points 196 

111.  Cusps 199 

112.  Conjugate  Points 201 

113.  Shooting  Points.  Stop  Points 203 

114.  Tracing  Curves 205 

Examples 206 

115.  Tracing  Polar  Curves.  Examples 210 


X 


CONTENTS. 


CHAPTER  XI. 

RADIUS  OF  CURVATURE,  EVOLUTES  AND  INVOLUTES — = 
ENVELOPES. 

AET.  PAGE 

116.  Curvature 216 

117.  Order  of  Contact  of  Curves 217 

118.  Dependence  of  Order  of  Contact  on  Arbitrary  Constants 218 

119.  Radius  of  Curvature.  Centre  of  Curvature 219 

120.  Second  Method 220 

121.  Radius  of  Curvature  in  Polar  Co-ordinates 222 

122.  Radius  of  Curvature  at  a Maximum  or  Minimum 222 

123.  Contact  of  Different  Orders 223 

Examples 224 

124.  Evolutes  and  Involutes 226 

125.  Equation  of  the  Evolute 228 

126  Normal  to  an  Involute  is  tangent  to  Evolute 230 

127.  Envelopes  of  Curves 231 

128.  Equation  of  the  Envelope  of  a Series  of  Curves 232 

Examples 233 


PART  II. 

SNTEGRAL  CALCULUS. 


CHAPTER  I. 

ELEMENTARY  FORMS  OF  INTEGRATION. 

129.  Definitions 238 

130.  Elementary  Rules  for  Integration 239 

131.  Fundamental  Forms 242 

132.  Integration  by  Transform&lion  into  Fundamental  Forms 243 

133.  Integrating  Factor.  Examples 247 

134.  Transposition  of  Variable  Factors.  Examples 249 

135.  Trigonometric  Reduction.  Examples 254 


CONTENTS. 


xi 


CHAPTER  II. 

INTEGRATION  OF  RATIONAL  FRACTIONS. 

ART. 

136.  Kational  Fractions. 

137.  Case  1.  Decomposition  of  a Rational  Fraction  

138  Case  2.  “ “ “ 

139.  Case  3.  “ « “ 

Examples. 

CHAPTER  III. 

INTEGRATION  OF  IRRATIONAL  FUNCTIONS  BT 
RATIONALIZATION. 

140.  Rationalization 

141.  Monomial  Surds 

142.  Binomial  Surds  of  the  First  Degree 

143.  Functions  of  the  Form  1 — 

(a  + bx2)* 

144.  Functions  containing  only  Trinomial  Surds 

145.  Binomial  Differentials 

r_ 

146.  Conditions  for  Rationalization  of  xm  ( a + bxn)i  dx  

Examples  

CHAPTER  IV. 

INTEGRATION  BY  SUCCESSIVE  REDUCTIONS. 

147.  Formula  of  Reduction 

148.  Formula  for  Diminishing  Exponent  of  x,  etc  

149.  Formula  for  Increasing  Exponent  of  x,  etc 

150.  Formula  for  Diminishing  Exponent  of  Parenthesis 

151.  Formula  for  Increasing  Exponent  of  Parenthesis 

Examples.  Applications  of  Formulae 

Logarithmic  Functions 

152.  Reduction  of  the  Form  j JE(Logx)ndx 

y'1  x'ndx 

n — - — 

log’1  x 


PAGE 

256 

256 

259 

262 

263 

269 

269 

270 

272 

272 

276 

277 

280 

285 

285 

287 

288 

289 

289 

295 

295 

297 


Xll 


CONTEXTS. 


AKT.  PAGE 

Exponential  Forms 299 

154.  Reduction  of  the  Form  J amIx"dx 299 

/(j:dx 

— — . 30C 

156.  Trigonometric  Functions . . 301 

157.  Formulae  of  Reduction  for  j sin”1  8 cos"  8 dO 303 

158.  Integration  of  sinm  6 cos"  8 dd 305 

159.  Reduction  of  the  Form  J ; xn  cos  ax  dx 307 

160.  Reduction  of  the  Form  &11  cos"  x dx. 308 

161.  Integration  of  f(x)  sin-1  x dx,  f (x)  tan-1  x dx,  etc 309 

/Jf) 

162.  Integration  of  dy  — 310 

a + b cos  8 

Examples 312 


CHAPTER  Y. 


INTEGRATION  BY  SERIES — SUCCESSIVE  INTEGRATION — IN- 
TEGRATION OF  FUNCTIONS  OF  TWO  VARIABLES. 
DEFINITE  INTEGRALS. 


163.  Integration  by  Series 

164.  Successive  Integration 

rn 

165.  To  Develop  the  nth  Integral  / X dxn  into  a Series. . 

166.  Integrations  of  Functions  of  Two  or  More  Variables 

(fill 

167.  Integration  of  = f(x,y) 

168.  Integration  of  Total  Differentials  of  the  First  Order. 

169.  Definite  Integrals.  Examples = 

170.  Change  of  Limits 

Examples 

Formulas  of  Integration 


319 

321 

323 

326 

326 

329 

331 

334 

337 

346 


CHAPTER  VI. 

LENGTHS  OF  CURVES. 

171.  Length  of  Plane  Curves  referred  to  Rectangular  Axes 347 

172.  Rectification  of  Parabola 348 


CONTEXTS. 


xiii 

tBT,  PAGE 

173.  Semi-cubical  Parabola 349 

174.  The  Circle 349 

175.  The  Ellipse 350 

176.  The  Cycloid 351 

177.  The  Catenary 352 

178.  The  Involute  of  a Circle 352 

179.  Rectification  in  Polar  Co-ordinates 353 

180.  The  Spiral  of  Archimedes 353 

181.  The  Cardioide 353 

182.  Length  of  Curves  in  Space 354 

183.  Intersection  of  Two  Cylinders.  Examples 355 

CHAPTER  VII. 

AREAS  OF  PLANE  CURVES. 

184.  Areas  of  Curves 360 

185.  Area  between  Two  Curves 361 

186.  Area  of  the  Circle 361 

187.  The  Parabola 362 

188.  The  Cycloid .362 

189.  The  Ellipse ; 363 

190  Area  bet  ween  Parabola  and  Circle . 363 

191.  Area  in  Polar  Co-ordinates 364 

192.  The  Spiral  of  Archimedes 364 

Examples 365 

CHAPTER  VIII. 

AREAS  OF  CURVED  SURFACES. 

193.  Surfaces  of  Revolution 369 

194.  Quadrature  of  the  Sphere 370 

195.  The  Paraboloid  of  Revolution 371 

196.  The  Prolate  Spheroid 372 

197.  The  Catenary 372 

198.  The  Surface  of  Revolution  generated  by  Cycloid 373 

199.  Surface  of  Revolution  in  Polar  Co-ordinates 374 

200.  The  Cardioide 374 

*01.  Any  Curved  Surfaces.  Double  Integration 375 

'{02.  Surface  of  the  Octant  of  a Sphere 376 

Examples 37 7 


CONTENTS. 


XiV 

CHAPTER  IX. 

VOLUMES  OF  SOLIDS. 

AST.  PAGE 

203.  Solids  cf  Revolution . 381 

204.  The  Sphere 381 

205.  Solid  of  Revolution  of  Cycloid 382 

206.  Solid  of  Revolution  generated  by  Cissoid 383 

207.  Volumes  of  Solids  bounded  by  any  Carved  Surface 383 

208.  Mixed  System  of  Co-ordinates 388 

209.  Cubature  in  Polar  Co-ordinates £89 

Examples. 390 


PART  I. 


DIFFERENTIAL  CALCULUS 


CHAPTER  I. 

FIRST  PRINCIPLES. 

1.  Constants  and  Variables.— -In  the  Calculus,  as  m 
Analytic  Geometry,  there  are  two  kinds  of  quantities  used, 
constants  and  variables. 

A constant  quantity,  or  simply  a constant,  is  one  whose 
value  does  not  change  in  the  same  discussion,  and  is  repre- 
sented by  one  of  the  leading  letters  of  the  alphabet. 

A variable  quantity,  or  simply  a variable,  is  one  which 
admits  of  an  infinite  number  of  values  within  certain  limits 
that  are  determined  by  the  nature  of  the  problem,  and  is 
represented  by  one  of  the  final  letters  of  the  alphabet. 

For  example,  in  the  equation  of  the  parabola, 

if  = 2 px, 

x and  y are  variables , as  they  represent  the  co-ordinates  of 
any  point  of  the  parabola,  and  so  may  have  an  indefinite 
number  of  different  values.  2p  is  a constant,  as  it  represents 
the  latns  rectum  of  the  parabola,  and  so  has  but  one  fixed 
value.  Any  given  number  is  constant. 

2.  Independent  and  Dependent  Variables.  — An 

independent  variable  is  one  to  which  any  arbitrary  value  may 


2 


FUNCTION  OF  ONE  OR  MORE  VARIABLES. 


be  assigned  at  pleasure.  A dependent  variable  is  one  whose 
value  varies  in  consequence  of  the  variation  of  the  inde- 
pendent variable  or  variables  with  which  it  is  connected. 

Thus,  in  the  equation  of  the  circle 

x2  + y2  = r”, 

if  we  assign  to  x any  arbitrary  value,  and  find  the  correspond- 
ing value  of  y,  we  make  x the  independent  variable,  and  y 
the  dependent  variable.  If  we  were  to  assign  to  y any  arbi- 
trary value,  and  find  the  corresponding  value  of  x,  we  would 
make  y the  independent  variable  and  x the  dependent 
variable. 

Frequently,  when  we  are  considering  two  or  more  varia- 
bles, it  is  in  our  power  to  make  whichever  we  please  the 
independent  variable.  But,  having  once  chosen  the  inde- 
pendent variable,  we  are  not  at  liberty  to  change  it  through- 
out our  operations,  unless  we  make  the  corresponding  trans- 
formations which  such  a change  would  require. 

3.  Functions. — One  quantity  is  called  a function  of 
another,  when  it  is  so  connected  with  it  that  no  change  can 
take  place  in  the  latter  without  producing  a corresponding 
change  in  the  former. 

For  example,  the  sine,  cosine,  tangent,  etc.,  of  an  angle 
are  said  to  be  functions  of  the  angle,  as  they  depend  upon 
the  angle  for  their  value.  Also,  the  area  of  a square  is  a 
function  of  its  side;  the  volume  of  a sphere  is  a function  of 
its  radius.  In  like  manner,  any  algebraic  expression  in  x,  as 

x3  — 2te2-f-  lx  -f-  c, 

is  a function  of  x.  Also,  we  may  have  a function  of  two  or 
more  variables : a rectangle  is  a function  of  its  two  sides : 
a parallelopiped  is  a function  of  its  three  edges  ; the  expres- 
sion tan  {ax  + by)  is  a function  of  two  variables,  x and  y; 
a?  + y1  + z2  is  a function  of  three  variables,  x,  y,  and  z : etc. 

When  we  wish  to  write  that  one  quantity  is  a function  of 


NOTA TION — GEOMETRIC  REPRESENTA TION. 


3 


one  or  more  others,  and  wish,  at  the  same  time,  to  indicate 
several  forms  of  functions  in  the  same  discussion,  we  use 
such  symbols  as  the  following: 

y — f(x) ; y = F{x) ; y = 0(*);  y =/'(*); 

V - f(x,  z) ; (p  (x,  y)  = 0 ; />,  y,  *)  = 0 ; 

which  are  read:  “ y equals  the  /function  of  a:;  ?/  equals  the 
large  A7  function  of  .r ; y equals  the  0 function  of  x ; y equals 
the /prime  function  of  x;  y equals  the  /function  of  x and 
z;  the  0 function  of  x and  y equals  zero;  the  / function  of 
x,  y,  and  z equals  zero;’'’  or  sometimes  “y  = f of  x, 
y = F of  x,”  etc.  If  we  do  not  care  to  state  precisely  the 
form  of  the  function,  we  may  read  the  above,  “ y — a func- 
tion of  x ; y — a function  of  x and  z ; a function  of  x and  y 
= 0 ; a function  of  x,  y,  and  z = 0.” 

For  example,  in  the  equation 

y ~ ax?  + lx  -f  c, 

y is  a function  of  x,  and  may  be  expressed,  y =f{x). 

Also,  the  equation 

ax 2 + Ixy  + cy%  = 0 

may  be  expressed,  / ( x , y)  = 0. 

In  like  manner,  the  equations 

y — ax?  -(-  bx~z  4-  cz3, 
and  y = ax 2 + bxz  4-  dz\ 

may  be  expressed,  y = f (x,  z)  and  y = (f>  (x,  z). 

Every  function  of  a single  variable  may  be  represented  geometri- 
cally by  the  ordinate  of  a curve  of  which  the  variable  is  the  cor- 
responding abscissa.  For  if  y be  any  function  of  x,  and  we  assigD 
any  value  to  x and  find  the  corresponding  value  of  y,  these  two  values 
may  be  regarded  as  the  co-ordinates  of  a point  which  may  be  con- 
structed. In  the  same  way,  any  number  of  values  maybe  assigned  to 
x,  and  the  corresponding  values  of  y found,  and  a series  of  points  con- 


4 ALGEBRAIC  AND  TRANSCENDENTAL  FUNCTIONS. 


strueted.  These  points  make  up  a curve  of  which  the  variable  ordi- 
nate  is  y and  the  corresponding  abscissa  is  x. 

In  like  manner  it  may  be  shown  that  a function  of  two  variables 
may  be  represented  geometrically  by  the  ordinate  of  a surface  of  which 
the  variables  are  the  corresponding  abscissas. 

4.  Algebraic  and  Transcendental  Functions. — An 

algebraic  function  is  one  in  which  the  only  operations  indi- 
cated are  addition,  subtraction,  multiplication,  division, 
involution,  and  evolution;  as. 


Transcendental  functions  are  those  which  involve  other 
operations,  and  are  subdivided  into  trigonometric,  circular, 
logarithmic,  and  exponential. 

A trigonometric  function  is  one  which  involves  sines,  tan- 
gents, cosines,  etc.,  as  variables.  For  example, 

y — sin  x\  y — tan2  x;  y = cos  x sec  x\  etc. 

A circular  function  is  one  in  which  the  concept  is  a 
variable  arc,  as  sin_1rc,*  cos-1  a:,  sec-1^,  cot-1  re,  etc.,  read, 
“ the  arc  whose  sine  is  x,  the  arc  whose  cosine  is  x,”  etc.  It 
is  the  inverse  of  the  trigonometric  function  ; thus,  from  the 
trigonometric  function,  y = sinrr,  we  obtain  the  circular 
function,  x = sin-1  y.  In  the  first  function  we  think  of  the 
right  line,  the  sine,  the  arc  being  given  to  tell  us  ichich  sine ; 
in  the  second  we  think  of  the  arc,  the  sine  being  given  to 
tell  us  which  arc.  The  circular  functions  are  often  called 
inverse  trigonometric  functions. 

* This  Dotation  was  suggested  by  the  use  of  the  negative  exponents  in  algebra. 
If  we  have  y = ax.  we  also  have  x = a 'y.  where  y is  a function  of  x,  and  x is  the 
corresponding  inverse  function  of  y.  It  may  be  worth  while  to  caution  the  begin- 
ner against  the  error  of  supposing  that  aim 1 y is  equivalent  to  — . — ; while  it  ii 


sin  y 


INCREASING  AND  DECREASING  FUNCTIONS. 


5 


A logarithmic  function  is  one  which  involves  logarithms 
Df  the  variables ; as, 

y — log  a:;  y = log  Va  — a; 

^ = 31og\/$T^’  etc‘ 

An  exponential  function  is  one  in  which  the  Y»riable 
enters  as  an  exponent ; as, 

y — ax\  y ~ xz  ; u = x^ ; etc. 

5.  Increasing  and  Decreasing  Functions. — An  in- 
creasing function  is  one  that  increases  when  its  variable 
increases,  and  decreases  when  its  variable  decreases. 

For  example,  in  the  equations 

y — ax3,  y = log  x,  y = V a2  + x>,  y = ax, 
y is  an.  increasing  function  of  x. 

A decreasing  function  is  one  that  decreases  when  its 
variable  increases,  and  increases  when  its  variable  decreases. 
Thus,  in  the  equations 

y = y = (a—  x)3>  y = log  -,  x2  + y2  = r», 

y is  a decreasing  function  of  x.  In  the  expression, 

y = (a  — xf, 

y is  a decreasing  function  for  all  values  of  x < a,  but  in- 
creasing for  all  values  > a.  In  the  expression 

y = sin  x, 

y is  an  increasing  function  for  all  values  of  x between  0° 
and  90°,  decreasing  for  all  values  of  x between  90°  and  270°, 
and  increasing  for  all  values  of  x between  270°  and  360°. 


6 


CONTINUOUS  FUNCTIONS. 


6.  Explicit  and  Implicit  Functions. — An  explicit 
function  is  one  whose  value  is  directly  expressed  in  terms  of 
the  variable  and  constants. 

For  example,  in  the  equations 

y = (a  — x)2,  y = V a2  — x2,  y = 'lax3  — 3z*, 

y is  an  explicit  function  of  x. 

An  implicit  function  is  one  whose  value  is  not  directly 
expressed  in  terms  of  the  variables  and  constants. 

For  example,  in  the  equations 

y3  — 3 axy  + x3  = 4,  x2  — 3 xy  + 2y  = 16, 

y is  an  implicit  function  of  x,  or  x is  an  implicit  function 
of  y.  If  we  solve  either  equation  with  respect  to  y,  we  shall 
have  y as  an  explicit  function  of  x ; also,  if  we  solve  for  x, 
we  shall  have  x as  an  explicit  function  of  y. 

7.  Continuous  Functions. — A function  of  x is  said  to 
be  a continuous  function  of  x,  between  the  limits  a and  b, 
when,  for  every  value  of  x between  these  limits,  the  cor- 
responding value  of  the  function  is  finite,  and  when  an 
infinitely  small  change  in  the  value  of  x produces  only  an 
infinitely  small  change  in  the  value  of  the  function.  If 
these  conditions  are  not  fulfilled,  the  function  is  discon- 
tinuous. 

For  example,  both  conditions  are  fulfilled  in  the  equations 

y — ax  -f  b,  y = sin  x, 

in  which,  as  x changes,  the  value  of  the  function  also 
changes,  but  changes  gradually  as  x changes  gradually,  and 
there  is  no  abrupt  passage  from  one  value  to  another;  if  x 
receives  a very  small  change,  the  corresponding  change  in 
the  function  of  x is  also  very  small. 

The  expression  V r 2 — x 2 is  a continuous  function  of  x 
for  all  values  of  x between  + r and  — r,  while  V x2  — r2 
is  discontinuous  between  the  same  limits. 


INFINITES  AND  INFINITESIMALS. 


1 


8.  Infinites  and  Infinitesimals. — An  infinite  quantity , 
or  an  infinite,  is  a quantity  which  is  greater  than  any  assign- 
able quantity. 

An  infinitesimal  is  a quantity  which  is  less  than  any 
assignable  quantity. 

An  infinite  is  not  tlie  largest  possible  quantity,  nor  is  an  infinitesi- 
mal the  smallest ; there  would,  in  this  case,  be  but  one  infinite  or 
infinitesimal.  Infinites  may  differ  from  each  other  and  from  a quan- 
tity which  transcends  every  assignable  quantity,  that  is,  from  absolute 
infinity.  So  may  infinitesimals  differ  from  each  other  and  from  abso- 
lute zero. 

The  terms  infinite  and  infinitesimal  are  not  applicable  to  quantities 
in  themselves  considered,  but  only  in  their  relation  to  each  other,  or  to 
a common  standard.  A magnitude  which  is  infinitely  great  in  com- 
parison with  a finite  magnitude  is  said  to  be  infinitely  great.  Also,  a 
magnitude  which  is  infinitely  small  in  comparison  with  a finite  mag- 
nitude is  said  to  be  infinitely  small.  Thus,  the  diameter  of  the  earth 
is  very  great  in  comparison  with  the  length  of  one  inch , but  very  small 
in  comparison  with  the  distance  of  the  earth  from  the  pole  star  ; and 
’t  would  accordingly  be  represented  by  a very  large  or  a very  small 
number,  according  to  which  of  these  distances  is  assumed  as  the  unit 
of  comparison. 

The  symbols  oo  and  0 are  used  to  represent  an  infinite 
and  an  infinitesimal  respectively,  the  relation  of  which  is 

oo  = ^ and  0 — — • 

U oo 

The  cipher  0 is  an  abbreviation  to  denote  an  indefinitely  small 
quantity,  or  an  infinitesimal— -that  is,  a quantity  which  is  less  than 
any  assignable  quantity — ana  does  not  mean  absolute  zero  ; neither 
does  oo  express  absolute  infinity. 

If  a represents  a finite  quantity,  and  x an  infinite,  then 

- is  an  Infinitesimal.  If  x is  an  infinitesimal  and  a is  finite, 
x 

- is  infinite ; that  is,  the  reciprocal  of  an  infinite  is  infini- 

00 

tesimal,  and  the  reciprocal  of  an  infinitesimal  is  infinite. 

A number  is  infinitely  great  in  comparison  with  another, 


8 


ORDERS  OF  INFINITES  AND  INFINITESIMALS. 


when  no  number  can  be  found  sufficiently  large  to  express  the 
ratio  betiveen  them.  Thus,  x is  infinitely  great  in  relation 
to  a,  when  no  number  can  be  found  large  enough  to  express 

X 

the  quotient  -•  Also,  a is  infinitely  small  in  relation  tc  x 
when  no  number  can  be  found  small  enough  to  express  the 
quotient  x and  - represent  an  infinite  and  an  infini- 
tesimal. 

One  million  in  comparison  with  one  miU/ionth  is  a very  large  num- 
ber, but  not  infinitely  large,  since  the  ratio  of  the  first  to  the  second 
can  be  expressed  in  figures : it  is  one  trillion  ; though  a very  large 
number,  it  is  finite.  So,  also,  one  millionth  in  comparison  with  one 
million  is  a very  small  number,  but  not  infinitely  small,  since  a num- 
ber can  be  found  small  enough  to  express  the  ratio  of  the  first  to  the 
second : it  is  one  trillionth,  and  therefore  finite. 


9.  Orders  of  Infinites  and  Infinitesimals. — But  even 

X 

though  - is  greater  than  any  quantity  to  which  we  can 

assign  a value,  we  may  suppose  another  quantity  as  large  in 
relation  to  tc  as  a;  is  in  relation  to  a : for,  whatever  the  mag- 
nitude of  x,  we  may  have  the  proportion 


a : x 


x 2 
x : 

a 


X^  • ... 

m which  — is  as  large  in  relation  to  £ as  a;  is  in  relation  to 

a 

ijfy 8 * t 

a , for  — will  contain  x as  many  times  as  x will  contain  a ; 
a 

X^ 

hence,  — may  be  regarded  as  an  infinite  of  the  second  order, 


X 

- being  an  infinite  of  the  first  order. 

Also,  even  though  ^ is  less  than  any  quantity  to  which 

we  can  assign  a value,  we  may  suppose  another  quantity  as 
small  in  relation  to  a as  a is  in  relation  to  x ; for  we  may 
have  the  proportion, 


ORDERS  OF  INFINITES  AND  INFINITESIMALS. 


9 


x : a : : a : — , 
x 


Co 

in  which  — is  as  small  in  relation  to  a as  a is  in  relation  to 

x 

a2  . . . 

x,  for  — is  contained  as  many  times  in  a as  a is  contained 

in  x ; hence,  — may  be  regarded  as  an  infinitesimal  of  the 


second  order,  - being  an  infinitesimal  of  th & first  order. 


We  may,  again,  suppose  quantities  infinitely  greater  and 

infinitely  less  than  these  just  named  ; and  so  on  indefinitely. 

Thus,  in  the  series 

, „ a a a , 

ax?,  ax?,  ax,  a,  -,  -5,  etc., 

X X?  X? 

if  we  suppose  a finite  and  x infinite,  it  is  clear  that  any 
term  is  infinitely  small  with  respect  to  the  one  that  imme- 
diately precedes  it,  and  infinitely  large  with  respect  to  the 
one  that  immediately  follows  it ; that  is,  ax?,  ax2,  ax  are 
infinites  of  the  third,  second,  and  first  orders,  respectively ; 

-,  (l-2,  ^ are  infinitesimals  of  the  first,  second,  and  third 

orders,  respectively,  while  a is  finite. 

If  two  quantities,  as  x and  y,  are  infinitesimals  of  the  first 
order,  their  product  is  an  infinitesimal  of  the  second  order ; 
for  we  have  the  proportion, 

1 : x : : y : xy. 

Hence,  if  x is  infinitely  small  in  relation  to  1,  xy  is  infinitely 
small  in  relation  to  y,  that  is,  it  is  an  infinitesimal  of  the 
second  order  when  x and  y are  infinitesimals  of  the  first 
order. 

Likewise,  the  product  of  two  infinites  of  the  first  order  is 
an  infinite  of  the  second  order. 

The  product  of  an  infinite  and  an  infinitesimal  of  the 
same  order  is  a finite  quantity.  The  product  of  an  infinite 


10 


RATIOS  OF  INFINITESIMALS. 


and  an  infinitesimal  of  different  orders  is  an  infinite  or  an 
infinitesimal,  according  as  the  order  of  the  infinite  is  higher 
or  lower  than  that  of  the  infinitesimal,  and  the  order  of  the 
product  is  the  sum  of  the  orders  of  the  factors. 

For  example,  in  the  expressions 

„ a a a a2 

ax?  x — k = a2,  ax2  x - — a2x,  ax  x —0  = 

x?  x x?  x2 

the  first  product  is  finite ; the  second  is  an  infinite  of  the 
first  order  ; the  third  is  an  infinitesimal  of  the  second  order. 

Though  two  quantities  are  each  infinitely  small,  they  may  have  any 
ratio  whatever. 

Thus,  if  a and  b are  finite  and  x is  infinite,  the  two  quantities 
^ and  ^ are  infinitesimals;  but  their  ratio  is  ^ , which  is  finite.  In- 
deed, two  very  small  quantities  may  have  a much  larger  ratio  than 
two  very  large  quantities,  for  the  value  of  a ratio  depends  on  the  rela- 
tive, and  not  on  the  absolute  magnitude  of  the  terms  of  the  ratio.  The 
ratio  of  the  fraction  one-millionth  to  one-ten-millionth  is  ten,  while  the 
ratio  of  one  million  to  ten  million  is  one-tenth.  The  latter  numbers  are 
respectively  a million  times  a million,  and  ten  million  times  ten  mil- 
lion, times  as  great  as  the  first,  and  yet  the  ratio  of  the  last  two  is 
only  one-hundredth  as  great  as  the  ratio  of  the  first  two. 

Assume  the  series 


in  which  the  first  fraction  is  one-millionth,  the  second  one-millionth 
of  the  first,  and  so  on.  Now  suppose  the  first  fraction  is  one-millionth 
of  an  inch  in  length,  which  may  be  regarded  as  a very  small  quantity 
of  the  first  order ; the  second,  being  one-millionth  of  the  first,  must 
be  regarded  as  a small  quantity  of  the  second  order,  and  so  on.  Now, 
if  we  continue  this  series  indefinitely,  it  is  dear  that  we  can  make  the 
terms  become  as  small  as  we  please  without  ever  reaching  absolute  zero. 
It  is  also  clear  that,  however  small  the  terms  of  this  series  become,  the 
ratio  of  any  term  to  the  one  that  immediately  follows  it  is  one  million. 

10.  Geometric  Illustration  of  Infinitesimals. — The 

following  geometric  results  will  help  to  illustrate  the  theory 
of  infinitesimals. 


GEOMETRIC  ILLUSTRATION  OF  INFINITESI3IALS.  11 


Let  A and  B be  two  points  on  the 
circumference  of  a circle  ; draw  the 
diameter  AE,  and  draw  EB  produced 
to  meet  the  tangent  AD  at  D.  Then, 
as  the  triangles  EAB  and  ADB  are 


similar,  we  have, 

BE 

AB 

(1) 

AE  — 

AD’ 

. AB 

BD 

(2) 

and  AE  = 

AD' 

E 


Now  suppose  the  point  B to  approach  the  point  A till  it 
becomes  infinitely  near  to  it,  then  BE  becomes  ultimately 
equal  to  AE;  but,  from  (1),  when 
BE  = AE, 

we  have  AB  = AD. 


Also,  becomes  infinitely  small,  that  is,  AB  becomes 

an  infinitely  small  quantity  in  comparison  with  AE.  Hence, 
from  (2),  BD  becomes  infinitely  small  in  comparison  with 
AD  or  AB  ; that  is,  when  AB  is  an  infinitesimal  of  the  first 
order,  BD  is  an  infinitesimal  of  the  second  order. 

Since  DE  — AE  < BD,  it  follows  that,  when  one  side  of 
a right-angled  triangle  is  regarded  as  an  infinitely  small 
quantity  of  the  first  order,  the  difference  between  the  hypoth- 
enuse  and  the  remaining  side  is  an  infinitely  small  quantity 
of  the  second  order. 


Draw  BN 
we  have, 


therefore, 


perpendicular  to  AD ; then,  since  AB  > AN, 

AD  - AB  < AD  - AN  < DN  ; 

AD  - AB  ^ DN  ^ AD 
BD  < BD  < DE' 


But  AD  is  infinitely  small  in  comparison  with  DE,  there- 
fore AD  — AB  is  infinitely  small  in  comparison  with  BD ; 


12 


AXIOMS. 


but  BD  is  an  infinitesimal  of  the  second  order  (see  above), 
hence  AD  — AB  is  an  infinitesimal  of  the  third  order. 

In  like  manner  it  may  be  shown  that  BD  — BN  is  an 
infinitesimal  of  the  fourth  order,  and  so  on.  [The  student 
who  wishes  further  illustration  is  referred  to  Williamson’s 
Dif.  Cal.,  p.  35,  from  which  this  wras  taken.] 

11.  Axioms. — From  the  nature  of  an  infinite  quantity, 
a finite  quantity  can  have  no  value  wThen  added  to  it,  and 
must  therefore  be  dropped. 

An  infinitesimal  can  have  no  value  -when  ad.ded  to  a finite 
quantity,  and  must  therefore  be  dropped. 

If  an  infinite  or  an  infinitesimal  be  multiplied  or  divided 
by  a finite  quantity,  its  order  is  not  changed. 

If  an  expression  involves  the  sum  or  difference  of  infinites 
of  different  orders,  its  value  is  equal  to  the  infinite  of  the 
highest  order,  and  all  the  others  can  have  no  value  when 
added  to  it,  and  must  be  dropped. 

If  an  expression  involves  the  sum  or  difference  of  infini- 
tesimals of  different  orders,  its  value  is  equal  to  the 
infinitesimal  of  the  lowest  order,  and  all  the  others  can  have 
no  value  when  added  to  it,  and  must  be  dropped. 

These  axioms  are  self-evident , and,  therefore,  axioms  in  the  strict 
sense.  For  example,  suppose  we  were  to  compare  the  mass  of  the  sun 
with  that  of  the  earth  : the  latter  weighs  about  six  sextillion  tons,  the 
former  weighs  about  355000  times  as  much.  If  a weight  of  one  grain 
were  added  to  or  subtracted  from  either,  it  would  not  affect  the  ratio 
appreciably  ; and  yet  the  grain,  compared  with  either,  is  finite — it  can 
be  expressed  in  figures,  though  on  the  verge  of  an  infinitesimal.  If 
we  divide  this  grain  into  a great  many  equal  parts — a sextillion,  for 
instance — and  add  one  of  these  parts  to  the  sun  or  the  earth,  the  error 
of  the  ratio  will  be  still  less  ; hence,  when  the  subdivision  is  continued 
indefinitely , it  is  evident  that  we  may  obtain  a fraction  less  than  any 
assiynable  quantity,  however  small,  which,  when  added  to  the  sun  or 
the  earth,  will  affect  the  above  ratio  by  a quantity  less  than  any  to 
which  we  can  assign  a value. 

By  reason  of  the  terms  that  may  be  omitted,  in  virtue  of  the  prin- 
ciples contained  in  these  axioms,  the  equations  formed  in  the  solution 


EXAMPLES. 


13 


of  a problem  will  be  greatly  simplified.  It  may  be  remarked  that  in 
the  method  of  limits ,*  when  exclusively  adopted,  it  is  usual  to  retain 
infinitely  small  quantities  of  higher  orders  until  the  end  of  the  calcu- 
lation, and  then  to  neglect  them  on  proceeding  to  the  limit ; while,  in 
the  infinitesimal  method,  such  quantities  are  neglected  from  the  be- 
ginning, from  the  knowledge  that  they  cannot  affect  the  final  result, 
as  they  necessarily  disappear  in  the  limit.  The  advantage  derived 
from  neglecting  these  quantities  will  be  evident  when  it  is  remem- 
bered how  much  the  difficulty  in  the  solution  of  a problem  is  increased 
when  it  is  necessary  to  introduce  into  its  equations  the  second,  third, 
and  in  general  the  higher  powers  of  the  quantities  to  be  considered. 


EXAMPLES. 


3X  I Q 

1.  Find  the  value  of  the  fraction  t— - ; — - , if  x is  infinite, 


and  a and  b finite. 


5x  -j-  b ' 


Since  a and  b are  finite,  they  have  no  value  in  comparison 

3 c 

with  x,  and  must  therefore  be  dropped,  giving  us  — = 
as  the  required  value  of  the  fraction. 


ox 


2.  Find  the  value  of  the  fraction 


2x 


3x  + b 


, if  x is  infini- 


tesimal, and  a and  b finite. 

Since  x is  an  infinitesimal,  it  has  no  value  in  comparison 
with  a and  b,  and  must  therefore  be  dropped,  giving  us  — ? 
for  the  required  value  of  the  fraction. 

3.  Find  the  value  of  — - , when  x is  infinite:  als# 

-4-  X 7 

when  x is  infinitesimal. 

Ans.  When  x is  infinite,  4;  when  infinitesimal,  2. 


, T-.  , ,,  n - ax?  + bx*  + cx  + e . 

4.  Find  the  value  of  — 5 s-1 1 — , when 

mx3  -f-  nx 2 +px-j-q 

infinite;  and  when  infinitesimal. 


x IB 


Ans.  When  x is  infinite,  - : when  infinitesimal,  -• 

m n 


* For  a discussion  of  limits,  see  Chapter  IIL 


14 


EXAMPLES. 


5.  Find  the  value  of  , when  x is  infinite: 

bxi  — AX  + 1 5 

and  when  infinitesimal. 

Ans.  When  x is  infinite,  oo  ; when  infinitesimal,  2. 

„ T-i-  i ii  , „ + 3a:2  + 2x  — 1 

6.  Find  the  value  ot  — — — , when  x is 

2a5  + Ac2  + 2x 

infinite;  and  when  infinitesimal. 

Ans.  When  x is  infinite,  0;  when  infinitesimal,  oo. 

*2  __  *7 nrrfirf 

7.  Find  the  value  of  — — 5 , when  x is  infinite ; and 

A.'Y'o  'YYl'T, 

when  infinitesimal. 

Ans.  Wrhen  x is  infinite,  0;  when  infinitesimal,  7 m. 


8.  Find  the  value  of 


z5 


2 a — x2 


, when  x is  infinite  ; and 


when  infinitesimal. 

Ans.  Wdien  x is  infinite,  oo;  when  infinitesimal,  0. 

9.  Find  the  value  of  when  x and  y are  infink 

tesimals.  ix  + S« 

Ans.  "We  do  not  know,  since  the  relation  between  x and  y 
is  unknown. 


CHAPTER  II 


DIFFERENTIATION  OF  ALGEBRAIC  AND  TRANSCEN- 
DENTAL FUNCTIONS. 

12.  Increments  and  Differentials. — If  any  variable, 
as  x,  be  supposed  to  receive  any  change,  such  change  is 
called  an  increment ; this  increment  of  x is  usually  denoted 
by  the  notation  Ax,  read  “difference  x,”  or  “delta  x, ” where 
A is  taken  as  an  abbreviation  of  the  word  difference.  If  the 
variable  is  increasing,  the  increment  is  + ; but  if  it  is 
decreasing,  the  increment  is  — . 

When  the  increment,  or  difference,  is  supposed  infinitely 
small,  or  an  infinitesimal,  it  is  called  a differential,  and  is 
represented  by  dx,  read  “ differential  x,”  where  d is  taken  as 
an  abbreviation  of  the  word  differential,  or  infinitely  small 
difference.  The  symbols  A and  d,  when  prefixed  to  a varia- 
ble or  function,  have  not  the  effect  of  multiplication  ; that 
is,  dx  is  not  d times  x,  and  Ax  is  not  A times  x,  but  their 
power  is  that  of  an  operation  performed  on  the  quantity  to 
which  they  are  prefixed. 

If  u be  a function  of  x,  and  x becomes  x + Ax,  the  cor- 
responding value  of  u is  represented  by  u + Aw. ; that  is,  the 
increment  of  u corresponding  to  a finite  increment  of  x is 
denoted  by  Am,  read  “ difference  n.” 

If  x becomes  x + dx,  the  corresponding  value  of  u is  rep- 
resented by  u-\-du  \ that  is,  the  infinitely  small  increment 
of  u caused  by  an  infinitely  small  increment  in  x,  on  which 
u depends,  is  denoted  by  du,  read  “ differential  u.”  Hence, 
dx  is  the  infinitesimal  increment  of  x,  or  the  infinitesimal 
quantity  by  which  x is  increased;  and  du  is  tic  correspond- 
ing infinitesimal  increment  of  u. 


16  CONSECUTIVE  POINTS — DIFFERENTIATION. 

The  differential  du  or  dx  is  + or  — according  as  the 
variable  is  increasing  or  decreasing,  i.  e.,  the  first  value  is 
always  to  be  taken  from  the  second. 

13.  Consecutive  Values.  — Consecutive  values  of  a 
function  or  variable  are  values  which  differ  from  each  other 
by  less  than  any  assignable  quantity. 

Consecutive  points  are  points  nearer  to  each  other  than 
any  assignable  distance. 

Thus,  if  two  points  were  one-millionth  of  an  inch  apart,  they  might 
be  considered  ‘practically  as  consecutive  points ; and  yet  we  might  have 
a million  points  between  them,  the  distance  between  any  two  of  which 
would  be  a millionth  of  a millionth  of  an  inch  ; and  so  we  might  have 
a million  points  between  any  twc  of  these  last  points,  and  so  on ; that 
is,  however  close  two  points  might  be  to  each  other,  we  could  still 
suppose  any  number  of  points  between  them. 

A differential  lias  been  defined  as  an  infinitely  small  in- 
crement, or  an  infinitesimal ; it  may  also  be  defined  as  the 
difference  between  two  consecutive  values  of  a variable  or 
function.  The  difference  is  always  found  by  taking  the  first 
value  from  the  second. 

In  the  Differential  Calculus,  we  investigate  the  relations 
between  the  infinitesimal  increments  of  variables  from  given 
relations  between  finite  values  of  those  variables. 

The  operation  of  finding  the  differential  of  a function  or 
a variable  is  called  differentiation. 

14.  Differentiation  of  the  Algebraic  Sum  of  a 
Number  of  Functions. 

Let  u = v + y — z,  (1) 

in  which  u,  v,  y,  z,  are  functions  of  x.* 


* We  might  also,  in  a similar  manner,  find  the  differential  of  a function  of  sev- 
eral variables  ; but  we  prefer  to  reserve  the  inquiry  into  the  differentials  of  functions 
of  several  variables  for  a later  chapter,  and  confine  ourselves  at  present  to  functions 
of  a single  variable. 


DIFFERENTIATION  OF  A PRODUCT.  17 

Give  to  x the  infinitesimal  increment  dx,  and  let  du,  dv, 
dy,  dz,  be  the  corresponding  infinitesimal  increments  of  u, 
v,  y,  z,  due  to  tbe  increment  which  x takes.  Then  (1) 
becomes 

u + du  = v + dv  + y + dy  — (z  + dz).  (2) 

Subtracting  (1)  from  (2),  we  have 

du  = dv  + dy  — dz,  (3) 

which  is  the  differential  required. 

Therefore,  the  differential  of  the  algebraic  sum  of 
any  number  of  functions  is  found  by  taking  the  alge- 
braic sum  of  their  differentials. 

15.  To  Differentiate 

y = ax  ± b.  (1) 

Give  to  x the  infinitesimal  increment  dx,  and  let  dy  be 
the  corresponding  infinitesimal  increment  of  y due  to  the 
ncrement  which  x takes.  Then  (1)  becomes 

y + dy  — a (x  + dx)  ± b.  (2) 

Subtracting  (1)  from  (2),  we  get 

dy  = adx,  (3) 

which  is  the  required  differential. 

Hence,  the  differential  of  the  product  of  a constant 
by  a variable  is  equal  to  the  constant  multiplied  by 
the  differential  of  the  variable;  also,  if  a constant  be 
connected  with  a variable  by  the  sign  + or  — , it  dis- 
appears in  differentiation. 

This  may  also  be  proved  geometrically  as  follows: 

Let  AB  (Fig.  2)  be  the  line  whose  equation  is  y = ax  + b, 
and  let  (x,  y)  be  any  point  P on  this  line.  Give  OM  (=  x) 


18 


GEOMETRIC  ILLUSTRATION. 


the  infinitesimal  increment  MM'  (=  dx),  then  the  cor< 
responding  increment  of  MP  (=  y) 
will  be  CP'  (=  dy).  Now  in  the  tri- 
angle CPP'  we  have 

CP'  = CP  tan  CPP' ; * 

or  letting  a = tan  CPP',  and  substi- 
tuting for  CP'  and  CP  their  values  dy 
and  dx,  we  have, 

dy  = adx. 

It  is  evident  that  the  constant  b will  disappear  in  differentiation, 
from  the  very  nature  of  constants,  which  do  not  admit  of  increase,  and 
therefore  can  take  no  increment. 

16.  Differentiation  of  the  Product  of  two  Func- 
tions. 

Let  u = yz,  (1) 

where  y and  z are  Loth  functions  of  x.  Give  x the  infini 
tesimal  increment  dx,  and  let  du,  dy,  dz  be  the  correspond- 
ing increments  of  u,  y,  and  z,  due  to  the  increment  which 
x takes.  Then  (1)  becomes 

u + du  = (y  -f-  dy)  ( z + dz) 

= yz  + zdy  + ydz  + dz  dy.  (2) 

Subtracting  (1)  from  (2),  and  omitting  dz  dy,  since  it  is 
an  infinitesimal  of  the  second  order,  and  added  to  others  of 
the  first  order  (Art.  11),  we  have 

du  = zdy  + ydz,  (3) 

which  is  the  required  differential. 

Hence,  the  differential  of  the  product  of  two  func- 
tions is  equal  to  the  first  into  the  differential  of  the 
second,  plus  the  second  into  the  differential  of  the 
first. 

* In  the  Calculus  as  in  the  Analytic  Geometry,  the  radius  is  always  regarded  a» 
\,  unless  otherwise  mentioned- 


GEOMETRIC  ILLUSTRATION. 


19 


This  may  also  be  proved  geometrically  as  follows : 

Let  2 and  y represent  the  lines  AB  b c ^ 

and  BO  respectively  ; then  will  u rep-  D HI  Ik 

resent  the  area  of  the  rectangle  ABCD. 

Give  AB  and  BC  the  infinitesimal  in- 
crements Bfl  (=  dz)  and  C c (=  dy) 
respectively.  Then  the  rectangle  ABCD  A 
will  be  increased  by  the  rectangles  B«C/i,  Fig>3i  B a " 

DC 'be,  and  C lied,  the  values  of  which 
are  ydz,  zdy,  and  dzdy  respectively  ; therefore 

du  — ydz  + zdy  + dz  dy. 

But  dzdy  being  an  infinitesimal  of  the  second  order  and 
connected  with  others  of  the  first  order,  must  be  dropped 
(Art.  11)  ; if  this  were  not  done,  infinitesimals  would  not 
be  what  they  are  (Art.  8)  ; the  very  fact  of  dropping  the 
term  dz  dy  implies  that  its  value,  as  compared  with  that  of 
ydz  + zdy  is  infinitely  small. 

The  statement  that  ydz  + zdy  + dzdy  is  rigorously  equal  to  ydz  + zdy 
is  not  true,  and  yet  by  taking  dz  and  dy  sufficiently  small,  the  error 
may  be  made  as  small  as  we  please. 

Or,  we  may  introduce  the  idea  of  motion,  and  consider 
that  dz  and  dy  represent  the  rate  at  which  AB  and  BC  are 
increasing  at  the  instant  they  are  equal  to  z and  y respec- 
tively. The  rate  at  which  the  rectangle  ABCD  is  enlarging 
at  this  instant  depends  upon  the  length  of  BC  and  the  rate 
at  which  it  is  moving  to  the  right  -f  the  length  of  DC  and 
the  rate  at  which  it  is  moving  upward.  If  we  let  dz  repre- 
sent the  rate  at  which  BC  is  moving  to  the  right,  and  dy  the 
rate  at  which  DC  is  moving  upward  at  the  instant  that 
AB  = z and  BC  = y,  we  shall  have  du  = zdy  -f-  ydz  as  the 
rate  at  which  the  rectangle  ABCD  is  enlarging  at  this  in 
stant.  (See  Price’s  Calculus,  vol.  i,  p.  41.) 


20 


DIFFERENTIATION  OF  A FRACTION. 


17.  Differentiation  of  the  Product  of  any  Num- 
ber of  Functions. 

Let  u — vyz.  (1) 

Then  giving  to  x the  infinitesimal  increments,  and  letting 
du,  dv,  dy,  dz  be  the  corresponding  increments  of  v,  v,  y,  z, 
(1)  becomes 

u + du  — (y  + dv)  (y  + dy)  ( z + dz).  (2) 

Subtracting  (1)  from  (2),  and  omitting  infinitesimals  of 
higher  orders  than  the  first,  we  have 

du  — yzdv  -f  vzdy  + vy  dz,  (3) 

and  so  on  for  any  number  of  functions. 

Hence,  the  differential  of  the  product  of  any  num- 
ber of  functions  is  equal  to  the  sum  of  the  products  of 
the  differential  of  each  into  the  product  of  all  the 
others. 

Cor. — Dividing  (3)  by  (1),  we  have 

du  dv  dy  dz 

u v y z x ' 

That  is,  if  the  differential  of  each  function  be  di- 
vided by  the  function  itself,  the  sum  of  the  quotients 
will  be  equal  to  the  differential  of  the  product  of  the 
functions  divided  by  the  product. 

18.  Differentiation  of  a Fraction. 

Let  u = -, 

y 

then  uy  = x',  \1) 

therefore,  by  Art.  16,  we  have 

udy  f ydu  = dx. 

Substituting  for  u its  value,  we  have 


DIFFER EXTI A TIOX  OF  A POWER. 


21 


-dy  + ydu  — dx. 


Solving  for  du,  we  get 


du  = 


ydx  — xdy 

yl  ; 


which  is  the  required  differential. 

Hence,  the  differential  of  a fraction  is  equal  to  the 
denominator  into  the  differential  of  the  numerator , 
minus  the  numerator  into  the  differential  of  the  de- 
nominator, divided  by  the  square  of  the  denominator. 


Cor.  1. — If  the  numerator  be  constant,  the  first  term  in 
the  differential  vanishes,  and  we  have 


du  = 


xdy 
V 2 ’ 


Hence,  the  differential  of  a fraction  with  a constant 
numerator  is  equal  to  minus  the  numerator  into  the 
differential  of  the  denominator  divided  by  the  square 
of  the  denominator . 


Cor.  2. — If  the  denominator  be  constant,  the  second  term 
vanishes,  and  we  have 


du  - 


dx 
V ’ 


which  is  the  same  result  we  would  get  by  applying  the  rule 
of  Art.  15. 


19.  Differentiation  of  any  Power  of  a Single  Va- 
riable. 

Let  y = xn. 

1st.  When  n is  a -positive  integer. 

Regarding  xn  as  the  product  x,  x,  x,  etc.,  of  n equal  fac- 
tors, each  equal  to  x , and  applying  the  rule  for  differentiating 
a product  (Art.  17),  we  get 


22  DIFFERENTIAL  OF  A POWER  OF  A VARIABLE. 


dy  — xn~l  dx  + xn~l  dx  + xn~x  dx  + etc.,  to  n terms. 

dy  = nx"~l  dx.  (1) 

2d.  When  n is  a positive  fraction. 


m 

Let  y — x'1; 

then  yn  = xm. 

Differentiating  this  as  just  shown,  we  have, 
nyn~x  dy  — mxm~l  dx. 


Therefore, 


, m xm~x  7 

X n yn-l 


m xm~x  y 

n yn 


dx 


m xn 


1 xn 


n 


dx  (since  = xm). 


771  i 

dy  = — X"  dx. 
y n 


3d.  When  n is  a negative  exponent,  integral  or 
fractional. 

Let  y - = x~n  ; 

then  y = — • 

v xn 


Differentiating  by  Art.  18,  Cor.  1,  we  have 

, nxn~x  dx  . , 

dy  — — — = — nx~n~x  dx.  (3) 

Combining  the  results  in  (1),  (2),  and  (3),  we  have  the 
following  rule:  The  differential  of  any  constant  power 
of  a variable  is  equal  to  the  product  of  the  exponent, 
the  variable  with  its  exponent  diminished  by  unity, 
and  the  differential  of  the  variable. 


EXAMPLES. 


Coe. — If  n — \,  we  have  from  (1), 

du  — \xl~l  dx  - ix~*  dx  = x ■ 
J ' “ 2Vz 


Hence,  the  differential  of  the  square  root  of  a varia ■ 
bis  is  equal  to  the  differential  of  the  variable  divided 
by  twice  the  square  root  of  the  variable. 


1.  Differentiate  y = 9 + 2x  + x3  -f  x2y3  — x7  + 2 axy. 

By  Art.  14,  we  differentiate  each  term  separately,  and 
take  the  algebraic  sum.  By  Art.  15,  the  constant  9 disap- 
pears in  the  differentiation  ; and  the  differential  of  2x  is  the 
constant  2,  multiplied  by  the  differential  of  the  variable  x, 
giving  2 dx.  By  Art.  19,  the  differential  of  x 2 is  2 xdx.  The 
term  x2y3  is  the  product  of  two  functions  ; therefore,  Art.  16, 
its  differential  is  x2d{y3)  -\-y3d(x2),  which,  Art.  19,  gives 
Zx2y2dy  + 2y3xdx.  In  like  manner  proceed  with  the 
other  terms,  giving  the  proper  rule  in  each  case.  The 
answer  is 

dy  — 2dx  + 3 x2  dx  -f  3 x2y2dy  + 2 xy3dx  — 7 x3dx 


EXAMPLES. 


-f-  2 ax  dy  -f  2 ay  dx. 


2.  u — ax?y2.  du  — 3ax2y 2 dx  -f-  2 ax?y  dy. 

3.  u = 2 ax  — 3x2  — ab&  — 7. 


4.  u = x2y^. 

5.  u = 2 bz~2  -f  3 ax§z^. 


du  = (2 a — 6x  — 4 abx?)  dx. 
du  — f x2y^  dy  + 2xif  dx. 


6.  u = x 


24 


EXAMPLES. 


YU  J>  Cl  , 1 

7.  u = ax 3 + - , — ox-. 

x2 

du  = ( \ax^  — ^ cZz. 

V ^ 2Vx/ 

8.  y2  = 2 px,  to  find  the  value  of  dy. 

dy  = - dx. 

y 


9.  a2y 2 + b2x2  = a2b 2,  to  find  the  value  of  dy. 

dy 

10.  x2  + y2  = r2,  to  find  the  value  of  dy. 


dy  =-^ix. 
a2y 


11.  u = 


b - 2 y2 
12.  u = (a  -f  bx  + cx2)5. 


dy=--a*. 
_ Aaydy  , 

(i  - W 


Regarding  the  quantity  within  the  parenthesis  as  a varia 
ble,  we  have,  by  Art.  19, 

du  — 5 (a  + bx  + cx2yd  (a  + bx  cx2) 

— 5 (a  + + cz2)4  (5  -f-  2cz)  Ja;. 


13. 


_ 2a;2  — 3 
4a;  + a*2 


By  Art.  18, 

(4a;  -f-  a;2)  d (2x2  — 3)  — (2x*  — 3)  <7  (4a;  + x2) 

du  - - (4 xTtfj2 

(4a-  + x2)  4a;  dx  — (2x2  — 3)  (4  -f-  2a;)  dx 
(4a;  + x2)2 
(8a;2  -f  6a;  + 12)  dx 
~ (4a;  + x2)2 


14.  u 


2a4 

a2  — x2 


du 


8 ah?  — ix>  , 
(a2  - x2)2  dX‘ 


EXAMPLES. 


25 


1 + X 

15-  y = T+& 

16.  y = ~r 


dy 


(1  2x  — x2)  clx 
~ (T~  + x2)2 


dy 


n 7 
— — r dx. 


17.  y = {ax2  — a?)4. 

dy  = 4 {ax2  — x B)3  (2 ax  — 3a?)  dx. 

18.  y = {a  + bz2)i.  dy  V (a  + bx2)  * bxdx. 

6aX—dx. 


19-  y — {&  + a?) 


dy  - (£2  + a«)< 


20.  y — Vx3  — a3  (Art.  19,  Cor.) 

dy 


d{x?  — a3)  3x2  dx 


21.  y — V %ax  — 

22.  y = —?==%' 

23.  y = Vox  + \/c2cfi. 


%Vx?  — a3  2 Va?  — a3 
(a  — x)  dx 
dy  V 2«a;  — x 2 


dy  - 


xdx 


(1  - a?)* 


ct~  -f"  3 cx  7 

dy  = --p~dx- 

2Vx 


. 7 (2 ax  + b)  dx 

24.  y ,=  Vax°  + bx  + c.  dy  = 

25.  y = (a?-f-  «)  (3a?  + &)  (Art.  16). 

dy  = {x3  + a)  d {3x2  + &)  + (3a;2  +5)  d (a:3  + a) 
= (15a?  + 3 bx2  + 6ax)  dx. 

26.  y — (1  + 2a?)  (1  + 4a?). 

dy  = 4a;  (1  + 3a;  + 10a?)  dx 

(a  + 3a;)  rfa; 

27.  y=  (a-x)Va  + x.  dy*--- 

{a— 3x)  dx 
dd  2 Va  — x 


28.  y = (a  + x)  Va  — x. 


26 


ILLUSTRATIVE  EXAMPLES. 


ILLUSTRATIVE  EXAMPLES. 


1.  In  the  parabola  y2  — 4x,  which  is  increasing  the  most 
rapidly  at  x — 3,  the  abscissa  or  the  ordinate  ? How  does 
the  relative  rate  of  change  vary  as  we  recede  from  the  ver- 
tex ? 

2 

Differentiating  y2  = \x,  we  get  dy  — -dx,  which  shows 

that  if  we  give  to  x the  infinitely  small  increment  dx,  the 

2 

corresponding  increment  of  y is  - times  as  great;  that  is, 
2 ^ 

the  ordinate  changes  - times  as  fast  as  the  abscissa.  At 

y_ 

x = 3,  we  have  y = Vl2.  Hence,  at  this  point, 


dy  = 


V 12  <Jl 


that  is,  the  ordinate  is  increasing  a little  over  one-half  as 
fast  as  the  abscissa  at  x — 3. 

At  x = 1,  y = 2,  and  dy  = dx;  that  is,  x and  y are 
increasing  equally;  in  general,  at  toe  focus  the  abscissa  and 
ordinate  of  a parabola  are  increasing  equally.  At  x — 4, 
y = 4,  and  dy  — \dx  ; that  is,  y is  increasing  £ as  fast  as 
x.  At  x — 9,  y — 6,  and  dy  — %dx ; that  is,  y is  increas- 
ing l as  fast  as  x.  At  x — 36,  y = 12,  and  dy  = \dx ; 
that  is,  y is  increasing  £ as  fast  as  x.  and  so  on.  We  see 
2 

from  the  equation  dy  = - (/z,  as  well  as  from  the  figure  of 

the  parabola,  that  the  larger  x becomes,  and  therefore  y,  the 
less  rapidly  y increases,  while  x continues  to  increase  uni- 
formly. 

2.  If  the  side  of  an  equilateral  triangle  is  increasing  uni- 
formly at  the  rate  of  an  inch  per  second,  at  what  rate  is 
its  altitude  increasing  ? Is  the  relative  rate  of  increase  ol 
the  side  and  altitude  constant  or  variable  ? 


ILLUSTRATIVE  examples. 


27 


Let  x = a side  of  the  triangle  and  y = its  altitude.  Then 

y'g 

y 2 = fa:2,  and  dy  — — dx,  which  shows  that  when  x takes 
the  infinitely  small  increment  dx,  the  corresponding  incre- 

y'g 

ment  of  y is  times  as  great;  that  is,  the  altitude  y 

always  changes  times  as  fast  as  the  side  x.  When  x is 
increasing  at  the  rate  of  ■§•  an  inch  per  second,  y is  increas- 
ing times  or  inches  per  second. 

Z 4: 


3.  A boy  is  running  on  a horizontal  plane  directly  towards 
the  foot  of  a tower  60  feet  in  height.  How  much  faster  is 
he  nearing  the  foot  than  the  top  of  the  tower  ? How  far  is 
he  from  the  foot  of  the  tower  when  he  is  approaching  it 
twice  as  fast  as  he  is  approaching  the  top?  When  he  is 
100  feet  from  the  foot  of  the  tower,  how  much  faster  is  he 
approaching  it  than  the  top  ? 

Let  x — the  boy’s  distance  from  the  foot  of  the  tower, 
and  y = his  distance  from  the  top.  Then  we  have 


y2  — x2  + &J 
dx  = ^ dy  ; 


V • 

that  is,  the  hoy  is  nearing  the  foot  - times  as  fast  as  he  is 
the  top. 


2d.  When  he  is  approaching  the  foot  of  the  tower  twice 
as  fast  as  he  is  the  top,  we  have  dx  = 2 dy,  which  in 


dx  — - d y 
x J 


gives  us  y — 2x,  and  this  in  y 2 = xz  + 604  gives  us 


3x2  = 602,  or 


60 

V73 


34.64. 


x 


28 


ILLUSTRATIVE  EXAMPLES. 


3d.  When  lie  is  100  feet  from  the  foot, 

y = j/lOO2  + 602  = 116.62, 

y 116.62  i • , ■ ; */  , 

and  - = — -rjr-  , which  m dx  = dy  gives 

dx  = 1.1662  dy  ; 

that  is,  he  is  approaching  the  foot  of  the  tower  1.1662  times 
as  fast  as  lie  is  the  top. 

4.  In  the  parabola  y 2 = 12x,  find  the  point  at  which  the 
ordinate  and  abscissa  are  increasing  equally;  also  the  point 
at  which  the  ordinate  is  increasing  half  as  fast  as  the 
abscissa.  Ans.  The  point  (3,  6);  and  the  point  (12,  12). 

5.  If  the  side  of  an  equilateral  triangle  is  increasing  uni- 

formly at  the  rate  of  2 inches  per  second,  at  what  rate  is  the 
altitude  increasing.  Ans.  VS  inches  per  second. 

6.  If  the  side  of  an  equilateral  triangle  is  increasing  uni- 
formly at  the  rate  of  5 inches  per  second,  at  what  rate  is  the 
area  increasing  when  the  side  is  10  feet  ? 

Ans.  f|V3  sq.  ft.  per  second. 

7.  A vessel  is  sailing  northwest  at  the  uniform  rate  of 

10  miles  per  hour ; at  what  rate  is  she  making  north  lati- 
tude ? Ans.  7.07+  miles  per  hour. 

8.  A boy  is  running  on  a horizontal  plane  directly  toward 
the  foot  of  a tower,  at  the  rate  of  5 miles  per  hour  ; at  what 
rate  is  he  approaching  the  top  of  the  tower  when  he  is  00 
feet  |rom  the  foot,  the  tower  being  80  feet  high  ? 

Ans.  3 miles  per  hour. 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS.  29 


LOGARITHMIC  AND  EXPONENTIAL  FUNC- 
TIONS. 


2Q  To  Differentiate  y — log-  x. — We  have 


( 


dx 


y + dy  = log  ( x + dx)  — log  x ( 1 + 

= log  Z + log  (l  + 

Subtracting,  we  have, 

7 7 dx\  (dx  dx  , \ 

dy  = log  ^1  + -)  = m {-  + etc./ 

(from  Algebra,  where  m is  the  modulus  of  the  system). 

dy  = d (log  x ) m ~ (Art.  11). 

This  result  may  also  be  obtained  as  follows: 

Let  y = ax.  (1) 

log  y — log  a -(-  log  x.  (2) 

By  Art.  15,  dy  — a dx,  (3) 

and  d (log  y)  — d (log  x).  (4) 

Dividing  (4)  by  (3),  we  get, 

d (log  y)  _ d (log  x) 
dy  a dx 


d (log  x)  ... 

— — — , from  (1); 


- dx 


d (log  y)  _ jy_ 

d (log  x)  dx 


30 


DIFFERENTIAL  OF  A LOGARITHM. 


Multiply  both  terms  of  the  second  fraction  by  the  arbi- 
trary factor  m,  and  we  have 

m dy 

d (i°g  y ) _ ~y~  , 

d( log  a:)  m dx'  l5' 

x 


We  may  suppose  m to  have  such  a value  as  to  make 


d (log  y) 

dy 
m — ; 

y 

(6) 

therefore, 

d (log  X)  — 

dx 
m — 

X 

(7) 

Similarly,  let 

y = 

Iz. 

•••  log  y = 

log  l + log  z. 

(8) 

Differentiatin 

g,  dy  = 

bdz, 

and 

d (log  y)  - 

d (log  z). 

Dividing  and 

substituting, 

dy 

d (log  y)  _ 

JL. 

d (log  z) 

dz 

z 

But 

d (log  y)  - 

du 
m — 

y 

d (log  z)  = 

dz 
m — 
z 

(9) 

In  the  same  way  we  may  show  that  the  differential  of  the 
logarithm  of  any  other  quantity  is  equal  to  m times  the 
differential  of  the  quantity  divided  by  the  quantity,  and 
hence  the  factor  m is  a constant,  provided  that  the  loga- 
rithms be  taken  in  each  case  in  the  same  system  ; of  course, 
if  the  logarithms  in  (8)  be  taken  in  a different  system  from 
those  in  (2),  the  numerical  values  of  logy  in  the  two  equar 


DIFFERENTIATION  01  AN  EXPONENTIAL  FUNCTION.  31 


tions  are  different,  and  therefore  the  m in  (ft)  is  different 
from  the  m in  (9).  Since  m is  a constant  in  the  same  sys- 
tem and  different  for  different  systems,  it  varies  with  the 
base  of  the  system,  as  the  only  other  quantities  involved  in 
logarithms  are  the  number  and  its  logarithm.  That  is,  m is 
a function  of  the  base ; its  value  will  be  computed  hereafter. 
(See  Rice  and  Johnson’s  Calculus,  p.  39;  also,  Olney’s  Cal- 
culus, p.  25.) 

Hence,  the  differential  of  the  logarithm  of  a quan- 
tity is  equal  to  the  modulus  of  the  system  into  the 
differential  of  the  quantity  divided  by  the  quantity. 

Cor. — If  the  logarithm  be  taken  in  the  Naperian*  system, 
the  modulus  is  unity,  and  we  have 


Hence,  the  differential  of  the  logarithm  of  a quan- 
tity in  the  JVaperian  system  is  equal  to  the  differen- 
tial of  the  quantity  divided  by  the  quantity. 

21.  To  Differentiate  y = ax. 

Passing  to  logarithms,  we  have, 

log  y = x log  a. 

Differentiating,  we  have 

m—=.dx  log  a ; 

y 

T y dx  log  a 

or  dy  = - — • 

^ m 

ax 

•\  dy  = d ( ax ) = — log  a dx. 


* So  called  from  the  name  of  the  inventor  of  logarithms;  also  sometimes  calleC 
natural  logarithms,  from  being  those  which  occur  first  in  the  investigation  of  p 
method  of  calculating  logarithms.  They  are  sometimes  called  hyperbolic  logarithms, 
from  having  been  originally  derived  from  the  hyperbola. 


32  DIFFERENTIATION  OF  AN  EXPONENTIAL  FUNCTION. 


Henc-e,  the  differential  of  an  exponential  function 
with  a constant  base  is  equal  to  the  function  into  the 
logarithm  of  the  base  into  the  differential  of  the  ex- 
ponent, divided  by  the  modulus. 

Cor.  1. — If  we  take  Naperian  logarithms,  we  have 
dy  — d ( ux ) = ax  log  a dx  (since  m = 1). 

Cor.  2. — If  a — c,  the  Naperian  base,  then 
log  a = log  e — 1, 
and  therefore  dy  = d (ex)  = exdx. 

Sch. — In  analytical  investigations,  the  logarithms  used 
are  almost  exclusively  Naperian,  the  base  of  which  system 
is  represen  ted  by  the  letter  e.  Since  the  form  of  the  differ- 
ential is  the  simplest  in  the  Naperian  system,  we  shall  in 
all  cases  understand  our  logarithms  to  be  Naperian,  unless 
otherwise  stated. 


22.  To  Differentiate  u = >jx. 

Passing  to  logarithms,  we  have, 

log  u — x log  y. 

Differentiating,  we  have, 

du  , , . dy 

— = log  y dx  + x — : 
u OJ  y 


or. 


du  = u log  y dx  + ux  — ; 

y 

du  = yx  log  y dx  + xyx~x  dy. 


Hence,  to  differentiate  an  exponential  function 
with  a variable  base,  differentiate  first  as  though  the 
base  were  constant  and  the  exponent  variable,  and 
second  as  though  the  base  were  variable  and  the  expo 
nent  constant,  and  take  the  sum  of  the  results. 


EXAMPLES. 


33 


EXAMPLES. 


1.  y = X log  X. 

2.  y = log  x2. 

3.  y = aloeX. 

4.  y — of. 

5.  y = a?1.  dy  = x? 


dy  = (log  x + 1)  dx 

, 2 dx 

dy  = — 
a x 

, _ a'°e  1 log  a dx 

ay  ~ 


dy  = a?  (F  log  a dx. 

1 


log  a;  (1  + log  x)  + 


xx  dx. 


6.  y = log  Vi  — d2. 

7.  y — log  (x  + Vl  + z2) 

8. 


, x dx 

d»  = ~ r- IS- 

dx 


ay  - Vl  + ** 

y = log  = log  (a  + x)  — log  (a  - x). 

2 adx 


dy  = 


a2  — x2 


§•  Jf  = log  = i l°g(l  +*)  -il°g(l—  *)- 


, dx 

= 1=7 


dy  = 


dx 


x log  x 


10.  y = log  (log  x). 

11.  y — log2  x. 

12.  y — xx. 

, ..  i V* + l—x 

13.  y = log  — === 

V x2  + 1 + x 

Multiplying  both  terms  by  the  numerator  to  rational^® 
the  denominator,  tve  get 


dy  = 2 logx^. 

JU 

dy  = x*  (log  x 4-  1 ) dx. 


34 


LOGARITHMIC  DIFFERENTIATION. 


y = log  [V x2  + 1 — z]\ 

7 2&; 

dy  = 


• V#2  4-  1 

14.  y = ex  (x  — 1). 

15.  y = e*  (a:2  — 2x  4-  2). 

16.  y = [• 

^ e*  + 1 

17.  y =:  e?  log  x. 

18.  y = e'og  *Wl*. 


dy  — 


dy  = eFxdx. 

dy  = (?3?dx. 
2e* 


dx. 


(c*  + 1)* 

dy  = e®  ^log  a:  4-  dx. 


Then 


log  y = log  V«2  4-  3s. 

y = 

dy  = 


y = V a2  4-  z3. 

xdx 


Va2  + 2s 


19.  y = 


<4 


1 + 3 

20.  y = log 


= 


xe*dx 
(1+2?' 
7 rfa:  rfa: 

* V22  + 1 


V#2 4- 1 4-  3 

21.  y = log  (ar  + «4- V2az+a-2).  dy  = ^==' 

22.  y = a4^~i.  <fy  = a\/—  1 z“  h-1 _ 1 dx. 


33.  y = 


a* 

a4 


23.  Logarithmic  Differentiation.  — When  the  function 
to  be  differentiated  consists  of  products  and  quotients,  the 
differentiation  is  often  performed  with  greater  facility  by 
first  passing  to  logarithms.  This  process  is  called  logarith* 
mic  differentiation. 


EXAMPLES. 


& 


EXAMPLES. 

1.  u = x [a2  + x2)  Va2  — x2. 

Passing  to  logarithms,  we  have 

log  u = log  x + log  (a2  + x1)  + | log  (a3  — x1\ 


du dx  2x  dx 

u x a2  -f-  x2 


xdx 


72  — .7-2 


du 


— I (a2-\-x2)Va2— x2  + 2x2Va2  — x2  — ^===-J 


dx 


w, 


, a4  + «2a;2  — 4.T4 

du  = — dx. 

V c2  — iC2 


^ _L 

2.  « = -5«  Passing  to  logarithms,  we  have 

X £v 

log  U — log  (1  + X1)  — log  (1  — X2). 

du  2x  dx  2x  dx  4a;  dx 

u 


8. 

L 


u — 
u — 


5.  u = 


• *=(!-*■)■ 
(a*  + 1)». 
a*  — 1 
ax  + l‘ 

Vl  + x 
Vl  — x 


2x  dx  2x  dx  

1 + x2  1 — x2  ~ (1  + x2)  (1  — £5) 

4a;  dx 


du  = 2ax(ax  + 1)  log  a dx. 
, 2 ax  log  a dx 

d*  = (J I iy  • 


du  — 


dx 


(1  - *)  Vl  - a* 


ILLUSTRATIVE  EXAMPLES. 

1.  Which  increases  the  more  rapidly,  a number  or  its 
logarithm?  How  much  more  rapidly  is  the  number  4238 
increasing  than  its  common  logarithm,  supposing  the  two 
to  be  increasing  uniformly?  While  the  number  increases 
by  1,  how  much  will  its  logarithm  increase,  supposing  the 


36 


ILLUSTRATIVE  EXAMPLES. 


latter  to  increase  uniformly  (which  it  does  not)  while  the 
camber  increases  uniformly. 

Let  x = the  number,  and  y its  logarithm ; then  wTe  have 

y = log  x; 


dy  = - dx, 
J x 


which  shows  that  if  we  give  to  the  number  ( x ) the  infinitely 
small  increment  (dx),  the  corresponding  increment  of  y is 

VI 

- times  as  great;  that  is,  the  logarithm  (y)  is  increasing 

171 

— times  as  fast  as  the  number.  Hence,  the  increase  in  the 
x 

common  logarithm  of  a number  is  >,  — , < the  increase 
of  the  number,  according  as  the  number  ( x ) <,  =,  > the 
modulus  (to). 

When  x = 4238,  we  have 


, m 7 .43429448  , 

dy  ~ 4238  dX  ~ 4238  dX * 

4238 

hence,  dx  = -43429443  dV  = about  9758  dy\ 

.43423448 

that  is,  the  increment  of  the  logarithm  is  - — — part  of 

the  increment  of  the  number,  and  the  number  is  increasing 
about  9758  times  as  fast  as  its  logarithm. 

While  the  number  increases  by  1,  its  logarithm  will  in- 
crease (supposing  it  to  increase  uniformly  with  the  number) 
43429448 

times  1 = .00010247;  that  is,  the  logarithm  of 

4238 

4239  would  be  .00010247  larger  than  the  logarithm  of  4238, 
if  it  ^’ere  increasing  uniformly,  while  the  number  increased 
from  4238  to  4239. 


Remark. — While  a number  is  increasing  uniformly,  its  logarithm 
is  increasing  more  and  more  slowly  ; this  is  evident  from  the  equation 

Tfl 

i y = — dx,  which  shows  that  if  the  number  receives  a very  small  in- 

X 


TRIGONOMETRIC  FUNCTIONS. 


om 

o • 


crement,  its  logarithm  receives  a very  small  increment ; but  on  giving 
to  the  number  a second  very  small  increment  equal  to  the  first,  the 
corresponding  increment  of  the  logarithm  is  a little  less  than  the  first, 
and  so  on  ; and  yet  the  supposition  that  the  relative  rate  of  change  of 
a number  and  its  logarithm  is  constant  for  comparatively  small  changes 
in  the  number  is  sufficiently  accurate  for  practical  purposes,  and  is  the 
assumption  made  in  using  the  tabular  difference  in  the  tables  of  loga 
rithms. 

2.  The  common  logarithm  of  327  is  2.514548.  What  is 
the  logarithm  of  327.12,  supposing  the  relative  rate  of 
change  of  the  number  and  its  logarithm  to  continue  uni- 
formly the  same  from  327  to  327.12  that  it  is  at  327  ? 

Ans.  2.514707. 

3.  Find  what  should  be  the  tabular  difference  in  the  table 

of  logarithms  for  numbers  between  4825  and  4826  ; in  other 
words,  find  the  increment  of  the  logarithm  while  the  num- 
ber increases  from  4825  to  4826.  Ans.  .0000900. 

4.  Find  what  should  be  the  tabular  difference  in  the  table 
of  logarithms  for  numbers  between  9651  and  9652. 

Ans.  .0000450. 

5.  Find  what  should  be  the  tabular  difference  in  the  table 
of  logarithms  for  numbers  between  7235  and  7236. 

Ans.  .0000601. 

TRIGONOMETRIC  FUNCTIONS. 

24.  To  Differentiate  y = sin  x.  (1) 

Give  to  x the  infinitely  small  increment  dx,  and  let  dy 
represent  the  corresponding  increment  of  y ; then  we  have 

y + dy  = sin  ( x + dx) 

= sin  x cos  dx  + cos  x sin  dx.  (2) 

Because  the  arc  dx  is  infinitely  small,  its  sine  is  equal  to 
the  arc  itself  and  its  cosine  equals  1 ; therefore  (2)  may  be 
written 


y + dy  = sin  x 4-  cos  x dx. 


(3) 


38 


TRIG  0N0METR1 C FUNCTIONS. 


Subtracting  (1)  from  (3),  we  have 

dy  = cos  x dx.  (4) 

Hence,  the  differential  of  the  sine  of  an  arc  is  equal 
to  the  cosine  of  the  arc  into  the  differential  of  the  a/  c 


25.  To  Differentiate  y — cos  x. 

Give  to  x the  infinitely  small  increment  dx,  and  we  havt 
y -f  dy  = cos  ( x + dx) 

= cos  x cos  dx  — sin  x sin  dx 
= cos  x — sin  x dx  (Art.  24). 

,\  dy  = — sin  x dx. 

Otherwise  thus: 

We  have  y = cos  x — sin  (90°  — x). 
Differentiating  hy  Art.  24,  we  have 

dy  — cos  (90°  — x)  d (90°  — x) 

= sin  x d (90°  — x). 
dy  — — sin  x dx. 

Hence,  the  differencial  of  the  cosine  of  an  arc  is 
negative  and  equal  to  the  sine  of  the  arc  into  the  dif- 
ferential of  the  arc.  (The  negative  sign  shows  that  the 
cosine  decreases  as  the  arc  increases.) 


26.  To  Differentiate  y — tail  x. 

We  have 


y = tan  x = 


COS  X 


Differentiating  by  Arts.  18,  24,  and  25,  we  have 
cos  x d sin  x — sin  x d cos  x 


dy  = 


cos2  X 

cos2  x + sin2  x 


cos2  x 
— see2  x dx. 


dx  = 


dx 


cos2  x 

dy  = sec2  x dx. 


TRIGONOMETRIC  FUNCTIONS. 


39 


Otherwise  thus: 

Give  to  x the  infinitesimal  increment  dx,  and  we  have 
y 4-  dy  — tan  (x  + dx) 

dy  — tan  ( x -f  dx)  — tana; 

tan  x + tan  dx 

— ; tan  x 

1 — tan  x tan  dx 

tan  x + dx  . , . . , , 

= ; , tan  x (since  tan  dx  = dx) 

1 — tan  x dx 

dx  + tan2  x dx  „ 

= — ; — = sec2  x dx 

1 — tan  x dx 

(since  tan  x dx,  being  an  infinitesimal,  may  be  dropped  from 
the  denominator). 

dy  — sec2  x dx. 

Hence,  the  differential  of  the  tangent  of  an  arc  is 
equal  to  the  square  of  the  secant  of  the  are  into  the 
differential  of  the  arc. 

27.  To  Differentiate  y = cot  x. 

We  have  y = cot  x = tan  (90°  — x). 

dy  = sec2  (90°  — x)  d (90°  — x). 
dy  — — cosec2  x dx. 

The  minus  sign  shows  that  the  cotangent  decre-ufitia  the  are 
Increases. 

Hence,  the  differential  of  the  cotangent  of  an  arc  is 
negative,  and  equal  to  the  square  of  the  cosecant  of 
the  arc  into  the  differential  of  the  arc. 

28.  To  Differentiate  y - sec  x. 

We  have  y =■  sec  x = — — • 

J cos  X 


40 


TRIGONOMETRIC  FUNCTIONS. 


dy  = — 


d cos  x 
cos2  x 


sm  x dx  • 

5 — = sec  x tan  x dx. 

cos2  x 


Hence,  the  differential  of  the  secant  of  an  arc  is 
equal  to  the  secant  of  the  same  arc,  into  the  tangent 
of  the  arc,  into  the  differential  of  the  arc. 

29.  To  Differentiate  y = cosec  x. 

We  have  y — cosec  x = sec  (90°  — x). 
dy  — d sec  (90°  — x) 

= sec  (90°  — x)  tan  (90°  — x)  d (90°  — x) 

— — cosec  x cot  x dx. 


Hence,  the  differential  of  the  cosecant  of  an  arc  is 
negative,  and  equal  to  the  cosecant  of  the  arc,  into  the 
cotangent  of  the  arc,  into  the  differential  of  the  arc. 


30.  To  Differentiate  y = vers  x. 

We  have  y = vers  x — 1 — cos  x. 

dy  = d (1  — cos  x)  = sin  x dx. 

Hence,  the  differential  of  the  versed-sine  of  an  arc 
is  equal  to  the  sine  of  the  arc  into  the  differential  of 
the  arc. 

31.  To  Differentiate  y = covers  x. 

We  have  y — covers  x = vers  (90°  — x). 

dy  = d vers  (90°  — x)  = sin  (90°  — x)  d (90°  — x) 
=z  — cos  x dx. . 

Hence,  the  differential  of  the  coversed-sine  of  an 
arc  is  negative,  and  equal  to  the  cosine  of  the  arc  into 
the  differential  of  the  arc. 


GEOMETRIC  DEMONSTRATION. 


41 


32.  Geometric  Demonstration. — The  results  arrived 
at  in  the  preceding  Articles 
admit  also  of  easy  demonstra- 
tion by  geometric  construction. 

Let  P and  Q be  two  consec- 
utive points*  in  the  arc  of  a 
circle  described  with  radius  = 1. 

Let  x — arc  AP  ; then 

dx  = arc  PQ. 

Prom  the  figure  we  have, 

PM  = sin  x\  NQ  = sin  (x  + dx)  ; 

•\  QR  = d sin  x. 

OM  = cos  x ; ON  ==  cos  ( x + dx)  ; 

*.  NM  = — d cos  x (minus  because  decreasing). 

AT  = tan  x ; AT'  = tan  ( x + dx)  ; 

TT'  = d tan  x. 

OT  = sec  x\  OT'  = sec  ( x + dx) ; 

.-.  DT'  — d sec  x. 

Now,  since  RP  and  QP  are  perpendicular  respectively  to 
MP  and  OP,  and  since  DT  and  TT'  are  also  perpendicular 
to  OT  and  OA  respectively,  the  two  infinitely  small  triangles 
PQR  and  DTT'  are  similar  to  MOP.  Hence  we  have  the 
following  equations : 

d sin  x — RQ  QP  cos  PQR 
= cos  x dx. 
d sin  x = cos  x dx. 


* All  that  is  meant  here  is  that  P and  Q are  to  be  reasoned  upon  as  though  they 
were  consecutive  points  ; of  course,  strictly  speaking,  consecutive  points  can  never 
be  represented  geometrically , since  their  distance  apart  is  less  than  any  assignable 
distance.  When  we  say  that  P and  Q are  consecutive  points,  we  may  regard  the 
distance  PQ  in  the  figure  as  representing  the  infinitesimal  distance  between  two 
consecutive  points,  highly  magnified. 


42 


GEOMETRIC  DEMONSTRATION. 


Also, 

and 


d cos  * = - PR  = - PQ  Em  PQR 
= — sin  x dx. 

d cos  x — — sin  x dx. 

d tan  x — TT'  = DT  sec  DTT'  DT  sec  x 
= OT-QP  sec  x (since  DT  - OT  QP 

- sec2  a;  dx. 

d tan  x — sec2  x dx. 

d sec  x = DT'  = DT  tan  DTT' 

= 0 T-QP  tan  x — sec  a:  tanr  dx. 
d sec  x — sec  x tan  x dx. 
cb  — — d (cot  x), 
cd  = — d (cosec  x). 


But  the  triangle  cbd  is  similar  to  the  triangle  OPM,  since 
cb  and  db  are  respectively  perpendicular  to  MP  and  OP. 
Hence  we  have 


d cot  x — — cb  = — db  cosec  deb 

— — bO-  QP  cosec  x — — cosec2  x dx. 

d cot  x — — cosec2  x dx. 


d cosec  x — — cd  — — db  cot  deb 
= — OJ-QP  cot  x 
— — cosec  x cot  x dx. 


d cosec  x — — cosec  x cot  x dx. 

From  the  figure  we  see  that  the  differential  of  the  versed* 
sine  is  the  same  numerically  as  that  of  the  cosine,  hut  with 
a contrary  sign,  i.  e.,  as  the  versed-sine  increases  the  cosine 
decreases  ; also  the  differential  of  the  coversed-sine  has  the 
same  value  numerically  as  that  of  the  sine. 


EXAMPLES. 


43 


1. 

2. 


3. 


4. 


5. 


6. 


7. 

8. 

9. 

10. 


11. 

12. 

13. 


14. 

15. 


EXAMPLES. 

y ~ sin  mx.  By  Art.  24  we  have, 

dy  — cos  mx  d ( mx ) — m cos  mx  dx. 
y - sin  ( x 2). 

dy  — cos  ( x 2)  d (re2)  = 2x  cos  (x2)  dx. 

y — sinm  x. 

dy  — m sinm_1  x d (sin  x)  = rn  sinm_1  x cos  x dx. 

y — COS3  X. 

dy  — 3 cos2  x d (cos  x)  = — 3 cos2  x sin  x dx 
= 3 (sin3  x — sin  x)  dx. 
y = sin  2x  cos  x. 

dy  = sin  2x  d cos  x + cos  x d sin  2x 

= — sin  2x  sin  x dx  + 2 cos  2x  cos  x dx. 


y — cot2  (x3). 

dy  = — 6x2  cot  x 3 cosec2  x3  dx. 

y = sin3  a;  cos#. 

dy  — sin2  x (3  — 4 sin2  x)  dx. 

y = 3 sin4  x. 

dy  — 12  sin3  a:  cosar  dx. 

y = sec2  5x. 
y — log  sin  x. 

dy  = 10  sec2  ox  tan  5x  dx. 

, d (sin  x ) 

dy  = — A 

J Sill  X 

(Art.  20)  = — — 1 dx  = cot  :r  dx. 
' sini; 

y — log  (sin2  x) 

= 2 log  sin  x.  dy  — 2 cot  x dx. 

y — log  cos  x. 

dy  — — tan  x dx. 

y — log  tan  x. 

-j  ' d tan  x 2 dx 

tan  x sin  2a; 

y — log  cot  x. 

, 2 dx 

dy  — — ■ — • 

sin  2x 

y = log  sec  x. 

dy  = tan  x dx. 

y — log  cosec  x. 

dy  = — cot  x dx. 

16. 


'*4 


ILL  USTRA  TI VE  EXAMPLES. 


,m  , /l  -f-  COS  X 

17.  y = — 

= £ log  (1  + cos  x)  log  (1  — cos  x). 

7 — dx 
“ ~ sin  x 

18.  y = e?  cos  x.  dy  = e?  d cos  x + cos  x de? 

— — e*  sin  x dx  -f  e®  cos  x dx 


= e* 

' (cos  a 

■ — sin 

x)  dx. 

19. 

y 

= x sin  x + cos  x. 

dy  = 

- X COS  X 

dx. 

20. 

y 

= xeC03*. 

dy  = 

= 

(1  — X 

' sin  x)  dx. 

21. 

y 

= eco“ 1 sin  x. 

dy  - 

_ ^COS  X 

(cos  X 

— sin2 * * * *  x) 

dx. 

22. 

y 

= log  V sin  x 

+ log  V7 cos  X 

= \ log  sin  x 

+ i log 

COS  X. 

dy  = 

\ (cot  X 

— tan 

x)  dx  ■■ 

- cot 2x 

dx. 

23. 

y 

= log  (cos  x + \/  — 1 

sin  x). 

dy  - 

= V-l 

dx. 

24. 

y 

— log  V7]  — 

sin  x 
sin  x 

dy  -- 

dx 

cos  X 

25. 

y 

- log  tan  (45 

0 + £*). 

dy  = 

dx 

cos  X 

26. 

y 

= sin  (log  x). 

dy  : 

= - cos  (log  x) 

dx. 

I LLUSTR.A 

lTIVE 

EX  A 

M P L ] 

ES. 

1.  Which  increases  faster,  the  arc  or  its  tangent?  When 

is  this  difference  least,  and  when  greatest  ? What  is  the 

value  of  the  arc  when  the  tangent  is  increasing  twice  as 

fast  as  the  arc,  and  when  increasing  four  times  as  fast  as  the 

arc  ? 

From  y — tan  x,  we  get  dy  = sec?  xdx,  which  shows 

that  if  we  give  to  the  arc  ( x ) the  infinitesimal  increment  dx, 


ILLUSTRATIVE  EXAMPLES. 


45 


the  corresponding  increment  of  the  tangent  ( y ) ia  sec2* 
times  as  great ; that  is,  the  tangent  (y)  is  increasing  secant 
square  times  as  fast  as  the  arc,  and  hence  is  generally  in* 
creasing  more  rapidly  than  the  arc.  When  * = 0,  sec  * = 1 ; 
therefore,  at  this  point,  the  tangent  and  the  arc  are  increas- 
ing at  the  same  rate.  When  * = 90°,  the  secant  is  infinite ; 
therefore,  at  this  point,  the  tangent  is  increasing  infinitely 
faster  than  the  arc. 

When  the  tangent  is  increasing  twice  as  fast  as  the  arc, 
we  have  dy  = 2dx,  or  sec2*  = 2,  which  gives  * = 45°; 
hence  at  45°  the  tangent  is  increasing  twice  as  fast  as  the 
arc. 

When  the  tangent  is  increasing  four  times  as  fast  as  the 
arc,  we  have  dy  = 4 dx,  or  sec2  * = 4,  which  gives  * = 60° ; 
hence  at  60°  the  tangent  is  increasing  four  times  as  fast  as 
the  arc. 

2.  Assuming  that  the  relative  rate  of  increase  of  the  sine, 
as  compared  with  the  arc,  remains  constantly  the  same  as 
at  60°,  how  much  does  the  sine  increase  when  the  arc  in- 
creases from  60°  to  60°  20'. 

Let  * = the  arc  and  y its  sine  ; then  we  have  y = sin  *, 

dy  = cos  * dx,  which  shows  that  the  increment  of  the 
sine  is  cosine  times  the  increment  of  the  arc.  Now  the  arc 

of  20'  =r  - .0058177  = dx\  therefore, 

180x3 

dy  — cos  60°  dx  = ^x. 0058176  = .0029088, 

which  is  the  increase  of  the  sine  on  the  above  supposition, 
and  is  a little  greater  than  the  increase  as  found  from  a table 
of  natural  sines,  as  it  should  be,  since  we  have  supposed  the 
sine  to  increase  uniformly  while  the  arc  was  increasing 
uniformly  from  60°  to  60°  20',  whereas  the  sine  is  increasing 
more  and  more  slowly  while  the  arc  is  increasing  uniformly. 
This  is  evident  from  the  equation  dy  — cos  * dx,  and  also 
from  geometric  considerations. 


46 


CIRCULAR  FUNCTIONS. 


3.  The  natural  cosine  of  5°  31'  is  .995368.  Assuming 

that  the  relative  rate  of  change  of  the  cosine  and  the  arc 
remains  the  same  as  at  this  point,  while  the  arc  increases  to 
5°  32',  what  is  the  cosine  of  5°  32'  ? Ans.  .995340. 

4.  The  logarithmic  sine  of  13°  49'  is  9.3780633.  Assum- 

ing that  the  relative  rate  of  change  of  the  logarithmic  sine 
and  the  arc  remains  the  same  as  at  this  point,  while  the  arc 
increases  to  13°  49'  10",  what  is  the  logarithmic  sine  of 
13°  49'  10"?  Ahs.  9.3781489. 

5.  The  log  cot  58°  2T  = 9.789863.  On  the  same  sup- 

position as  above,  what  is  the  decrease  of  this  logarithm  for 
1 second  increase  of  arc.  Ans.  .00000471. 

6.  A wheel  is  revolving  in  a vertical  plane  about  a fixed 
centre.  At  what  rate,  as  compared  with  its  angular  velocity, 
is  a point  in  its  circumference  ascending,  when  it  is  60° 
above  the  horizontal  plane  through  the  centre  of  motion. 

Ans.  Half  as  fast 

CIRCULAR  FUNCTIONS, 

33.  To  Differentiate  y = sin-1  x.* 

We  have  x — sin  y ; 

therefore,  dx  = cos  y dy  — V (1  — sin2  y)  dy 

= \/l  — o?  dy. 

dx 

•••  dy  = vct  = d (sin_1  *)• 

34.  To  Differentiate  y = cos-1  x. 

We  have,  x = cos  y ; 

therefore,  dx  — — sin  y dy  — — yl  — cosh/  dy 

- - — -y/l  — dy. 

■■■  dy= = i (cos->  x). 

yl  — x 2 


* This  notation,  as  already  explained,  means  y = the  arc  whose  sine  is  x. 


CIRCULAR  FUNCTIONS. 


47 


35.  To  Differentiate  y — tan-1  x. 

We  have  x = tan  y ; 

therefore,  dx  = sec ^y  dy  = (1  + tan2  y)dy 

— (1  + x 2)  dy. 

dV  = j f*  = d (tan-1  x). 

36.  To  Differentiate  y = cot-1  x. 

W e have  x = cot  y ; 

therefore,  dx  = — cosec 2y  dy  — — (1  + cot2  y)  dy 

= — (1  + z2)  %• 

dy  — - = d (cot-1  a). 


37.  To  Differentiate  y = sec-1  x. 

We  have  x = sec  y ; 

therefore,  dx  = sec  y tan  y dy  — sec  y V sec2  y — 1 dy 

= a:  V x2  — 1 dy. 


dy  = 


<?a; 


x*Jx2  — 1 


— (sec-1  a;). 


38.  To  Differentiate  y = cosec-1  a?. 

We  have  a;  = cosec  y ; 

therefore,  dx  = — cosec  y cot  y dy 

= — cosec  y V cosec2  y — 1 dy 


= — x's/  x 2 — 1 dy. 


1-8 


EXAMPLES. 


39.  To  Differentiate  y = vers-1  X • 

We  have  z = vers  y ; 

therefore,  dx  = sin  y dy  — Vl  — cos2  y dy 

= Vl  — (1  — vers  y)2.dy 
= V%  vers  y — vers2  y dy 
= V%x  — x3  dy. 

dy  ■—  ■ ^X  - -z.  — d (vers-1  z) 

V 2z  — z2 

40.  To  Differentiate  y — covers-1  x. 

We  have  x — covers  y\ 

therefore,  dx  — — cosy  dy  = — Vl  — sin’ y dy 
= — Vl  — (1  — covers  y)2  dy 
= — V%  covers  y — covers2  y dy 

= — a/2z  — z2  dy. 
dx 

dy  = : — : = r?  (covers-1  z). 

V%x  — z2 


EXAMPLES. 


1.  Differentiate  y = sin-1  -• 

3 a 

We  have,  by  Art.  33, 

,x 


d- 


dy  = 


a 


dx 

a 


V'-* 


dx 

V cd  — »• 


2.  Differentiate  y'  — a sin-1  — 

3 a 


EXAMPLES. 


4S 


We  have,  Art.  33,  dy'  — 


dx 
a — - 
a 


adx 


1 — 


^ V a2  — x2 


Geometric  illustration  of  Examples  1 and  2 : 

Let  OA  - 1,  OA'  — a,  y — arc  AB,  y'  - arc  A'B', 
x = M'B'. 

B'M'  _ x 
OB'  ~ a’ 

arc  AB  = sin-1  BM 
B'M' 


Now 


= sm' 


-l 


= sm 


OB' 

x 

a 


= y (see  Ex.  1). 

A'B'  = A'O  • arc  AB  = A'O  • sin"1 


B'M' 

OB' 


= a sin-1  - = y’  (see  Ex.  2). 

Also,  A'B'  =r  sin-1  B'M'  = sin-1  x (to  radius  a) 

X 

.*.  a siu-1  - (to  radius  1)  = sin-1  x (to  radius  a). 


Hence,  in  Example  1,  y is  the  arc  AB  (to  radius  1),  and 

X 

is  given  in  terms  of  the  sine  - (to  radius  1)  ; while  in 
Example  2,  y'  is  the  arc  A'B'  (to  radius  a),  and  is  given  in 

X 

terms  of  the  sin  - (to  radius  1). 

a ' 

If  we  give  B'M'  (which  is  x in  both  examples)  an  incre- 
ment (—  dx),  the  corresponding  increment  in  A'B'  will  be 
a times  as  great  as  that  on  AB ; that  is,  dy'  in  Ex.  2 is  a 
times  dy  in  Ex.  1. 


50 


EXAMPLES. 


y = cos-1  -• 

dy  = 

dx 

V a?  — x1 
adx 

y — tan-1  -• 

dy  = 

a2  + a:2 

y = cot-1  -• 
^ a 

dy  — 

adx 

d 1 + a;2 

y — sec  1 — 
27  a 

dy  = 

adx 

x\/  x1  — ad 

X 

y — cosec-1  — 
a 

dy  = 

adx 

x V a;2  — a2 

X 

y — vers-1  — 
v a 

dy  = 

dx 

V 2aa;  — x2 

X 

y = covers-1 — 
a 

dy  — 

dx 

V 2aa;  — a:2 

10.  y — a cos 


dy  = 


adx 

a 


adx 


\A-I 


Vo2  — 


11.  i — a tan  1 -•  dy  = 
^ a 


dx 

a 


_ a2(7a; 

a:2  a2  -+•  z2 
1 + a2 


<7a: 


IS.  y — a cot-1  -•  dy  = — 

5 O'7 


1 + 


a2dx 
a2  + ofi 


dx 


IS.  y —■  a sec-1  dy  = 
a J 


a?dx 


x /x2 

a V a2 


ay  a;2 


MISCELLANEOUS  EXAMPLES. 


51 


14. 


x 

a cosec  1 — 
a 


dx 


a2dx 


Xi\/  x 2 — a 2 


15. 

16. 


y = a vers-1  -•  dy 


y 


adx 


V 2 ax  — a? 


adx 

V — x? 


MISCELLANEOUS  EXAMPLES. 


, a -f-  x , 3 a — x , 

1.  y — -------  ■ dy  = 7 dx . 

V a — x 2 (a  — x)v 

2.  y = y x — V a?  — a?. 

■j  _ (a;  + \/«2  — a?) 

2 V«2  — a?  (a;  — Va2  — a?)£ 

n ? _ X 

x + ^/i  — a;2  y 2x  ( 1 — a;2)  + y 1 — #2 


4.  y = 


a? 

\/(T—  a?)3’ 


= 


3a?  (fa; 

(1  _a«)S* 


5.  y = 


a4 


^ V a2a2  — a? 


dy 


— a4  («2  — 2a:2)  (fa; 
2a?  (a2  — a?)* 


52 


MISCELLANEOUS  EXAMPLES. 


Vx2  + 1 — X 

6.  y — — - — 

Vx2  + 1 + X 

In  fractions  of  this  form,  the  student  'will  find  it  an  ad- 
vantage to  rationalize  the  denominator,  by  multiplying  both 
terms  of  the  fractiou  by  the  complementary  surd  form  ; that 
is,  in  the  present  case,  by  Vx2  + 1 — x.  Thus, 


Vx2  4-  1 — x Vx2  +~i  — x 

y ~ — x —z 

V x2  + 1 + X V*2+  1 — X 

_ {Vx2  + 1 — x)2 


••  iy  = *Wx'+1-r)(Vm~l) 

= 2 Ux  - dx. 

\ Vx2  + l } 


dx 


'•  »-Vi 


1 — Vx  _ V\  — Vx  _ Vi  — x 
+ Vx  V i + Vx  1 + Vx 

dx 


dy  = 


2(l  + Vx)  Vx  — x2 


e ■ y = 


Vl  + x + Vl  — x 
Vi  +x  — Vi  — x 


1 + Vl-x2^ 

d»  = - 


i.  y = Vx  ■ y Vx  + 1. 

dy  = 


7Vz  + 4 

1 2V^-y/  \V;  + 1 


: dx. 


MISCELLANEOUS  EXAMPLES. 


53 


10.  y = 


a — + V{c2  — a?)2 


35  4 £ 

^ 2xV  x V(c2  — a:2)  ^ 

4 \ / a + v"  (c2  — a;2)2 

V v x 


11.  y — 2x — 1 — -j/7 2.t — 1 — V2x — 1 — etc.,  ad  inf 

Squaring,  we  have, 


if  = 2x — 1 — y 2x — 1 — |/ 2x — l — V 2x—  1 — etc.,  ad  inf. 
Hence,  y2  — 2x  — 1 — y ; 


ana 


y — — I ± i V 82;  — 3. 
2 dx 


•*.  dy  = ± 


12.  y = 


V8x  — 3 
Vl  -fa:2  + V 1 — a;2 


VT+  a;2  — V 1 — a2 

* = -*(1  + vrM*t 


13.  y = log  — 

Vl  + ^ 


dy  - 


dx 


x (1  + X2) 


„ . , Vl  + x + Vl  — x , dx 

14.  y = log  , dy  = • 

Vl  + x — Vl  — X XVl — 2s 


15.  y = log 


V x2  + a2  — 


, dx  dx 

dy  = — + 


Vx2  + 


54 


MISCELLANEOUS  EXAMPLES. 


16. 


17. 


18. 

19. 

20. 
21. 

22. 

23. 

24. 


25. 


26. 

27. 

28. 
29. 


y = log  \_v i + x2  + vi  — z2] 

dy 

a ( 2x  — a) 


, _ dx  dx 

y ~ ~ 


y — log  (x  — a) 

y =z  ax\ 
y — eF  (1  — z3). 


_ ex  — e-x 

y — ex  + e~x‘ 

y = log  (e®  + e~x). 
y = of. 


nrl  _L_  n% 

, dy  = - — cfo. 

(x  — a)2  3 (x — a)3 

dy  — 2 a*2  log  a xdx. 

dy  = ex  (1  — 3a:2  — a:3)  dx. 

4 dx 


dy  = 


(e?  + e"*)2 


ex_e-x 

= x. ~dx. 

J ex  + e~x 


dy  = 


_ X?  (1  — log  x) 


y = 2e  (a:^  — 3a:  + 6ar*  — 6).  dy  = xe^*  dx. 


dx. 


y = 


(a:-l)f 


(x  — 2)i  (x  - 3)* 


(See  Art.  23.) 


(.-!).  (7^ + 30.-97)^ 
12  (x  — 2)1  (x  — 3)  ^ 

_ (.r  + l)Hr  + 3)t  (Art  23  x 

y - {x  + 2y  (Art.  23.) 

. x2  (x  + 3)*  dx 

dy  = 1 r* 

(x  + 2)3  (x  + l)t 


x Vx’—l 


y — 


dy  — 


2(a/ x2  — l)  e: 


X Vx- — 1 


dx 


X + V x2  — 1 
y — sin  x — £ sin3  x. 
y = \ tan3  x — tan  x -\-  x. 
y = -J  tan3  x + tan  x. 


x + Vr2  — l 
dy  — cos3  a:  dx. 
dy  — tan4  x dx, 
dy  — sec4  a:  dx. 


MISCELLANEOUS  EXAMPLES. 


55 


30.  y 

31.  y 

32.  y 

33.  y 

34.  y 

35.  y 

36.  y 

37.  y = 

38.  y = 

39.  y = 

We  have, 

40.  = 
We  have. 


sin  ex. 

tan2  x -f  log  (cos2  x) 
log  (tan  x + sec  x). 
sin  x 


dy  = ex  cos  e*  dx. 
dy  — 2 tan3  z dx. 
dy  — sec  x dx. 


1 + tan  x 


(cos3  x — sin3  x)  j 

y ~ (sin  x + cos  x )2  Xm 


log 


sjl 


cos  x — b sin  x 


cos  x + b sin  x 
dy  — 


— ab  dx 


a2  cos2  x — b2  sin2  x 


j. 

tan  e*. 


dy  — 


tan  Vl 

— X. 

dy 

^.sin  x 

dy  : 

= afinr  ( 

2 

3 cos  x 

sin2  x cos  x 

sin2  x 

sin-1  — 

X 

i 1 

ez  sec2  ex  dx 
~ V 

(sec2  VI  — x)dx 
2Vl-rx 


+ 3 log  tan  -• 
dy  = 


2 dx 


sin3  a;  cos2 a? 


Vl  + x2 

dx  1 


+ x? 

dx 


(1  a X2)i  ' (1  + X2)i  1 + X? 

cos-1  a:  Vl  — x2  = cos-1  V x2  — x4. 

dy  = — d V x2  — xi  -i-  Vl  — (x2  — x4) 
( 1 — 2z2)  dx 


Vl  — xs 


4-  Vl  — x2  + X*. 


56 


MISCELLANEOUS  EXAMPLES. 


dy  = 


(1  — 2x2)  dx 
V(l  — x2)  (1  — x2  + xi) 


2dx 


41.  y = sin"1  (2a;  VI  — a;2).  dy  = 


42.  y = sin-1  (3a;  — 4aA). 

2a; 


dy  = — - 


3c/ar 


43.  y = tan-1 

44.  y — sin-1 


1 — z2 

1 — x2 
1 + a; 2 


, 2dx 

dy  = rr^' 


Vl  — a* 

Idj 
+" 


7 2rfa; 

^ = -1+5' 


45.  a:  = vers-1  y — V 2ry  — y2  (where  vers_1y  is  taken 
so  radius  r). 

We  may  write  this  (see  Art.  40,  Exs.  1 and  2), 
x = r vers-1  - — V 2 vy  — y3. 


dx  — 


rdy 


rdy  — ydy 
V2ry  — y3  V2ry  — y2 

ydy 

V 2ry  — y2 


46.  y = a;  V a2 — a;2+<z2  sin-1  -•  dy  = 2 Va2—x2dx. 


. 4/1  4-  a;  , dx 

47.  y = log  \Y  x—  + i tan  1 as.  rfy  = 


2a; 


48.  y = vers1  ^ 

49.  y — e1’"-1*. 


= 


dx 


dy  = ’* 


V9a;  — a;v 
dx 


l + ai* 


MIS  CEL  LANEO  US  EXAMPLES. 


57 


50. 


51. 

52. 


y xsln~lx . 

dy  — afin_1: 


y — sec-1  nx. 


y = siu  1 


V a2  + 


~x  log  x + (1  — x2)-*  sin  1x~ 
x (1  — x2)i 

, dx 

dy  — 

=•  dy  — 


dx. 


XV  n2x2  — 1 
adx 
a 2 + x2 


53. 

54. 


y = sin-1 
y - - tan-1 


V si 


n a". 


cos  x 


1 + COS  X 


dy  = (v"  1 + cosec  x)  tfx. 

dy  — \dx. 


55.  y = 


x sm_1  x 
Vl  — x2 


sin  1 x 


+ log  V I - x2.  dy  = -^—-dx, 
8 (l-x2)i 


56.  y = (x  -\-  a)  tan-1  — V «x. 

dy  = tan-1  'Y' ~ dx. 

x\/h 


57.  y = sec-1 


2\/x2  + x — 1 


dy  = 


dx 

XV  x2+~x  — 1 


58. 


?/  = tan  1 


3a2x  — x3 
«3  — 3ax2 


dy  = 


Zadx 
a 2 + x2 


59. 


y - sin-1 


x-\/ « — & 

Vx  (1  + x2) 


dy  = 


A /a  — l 

(1  + X2)  y'fl  _J_ 


dx. 


60.  If  two  bodies  start  together  from  the  extremity  of  the 
diameter  of  a circle,  the  one  moving  uniformly  along  the 
diameter  at  the  rate  of  10  feet  per  second,  and  the  other  in 
the  circumference  with  a variable  velocity  so  as  to  keep  it 
always  directly  above  the  former,  what  is  its  velocity  in  the 


58 


MISCELLANEOUS  EXAMPLES. 


circumference  when  passing  the  sixtieth  degree  from  the 

starting-point?  . 20  „ . , 

° r A ns.  — — feet  per  second. 

V3 

61.  If  two  bodies  start  together  from  the  extremity  of  the 
diameter  of  a circle,  the  one  moving  uniformly  along  the 
tangent  at  the  rate  of  10  feet  per  second,  and  the  other  in 
the  circumference  with  a variable  velocity,  so  as  to  be  always 
in  the  right  line  joining  the  first  body  with  the  centre  of 
the  circle,  what  is  its  velocity  when  passing  the  forty-fifth 
degree  from  the  starting-point.  Ans.  5 ft,  per  second. 


CHAPTER  III 


LIMITS  AND  DERIVED  FUNCTIONS. 

41.  Limiting  Values. — The  rules  for  differentiation 
have  been  deduced,  in  Chapter  II,  in  accordance  with  the 
method  of  infinitesimals  explained  in  Chapter  I.  We  shall 
now  deduce  these  rules  by  the  method  of  limits. 

The  limit,  or  limiting  value  of  a function,  is  that  value 
toward  which  the  function  continually  approaches,  till  it 
differs  from  it  by  less  than  any  assignable  quantity,  while 
the  independent  variable  approaches  some  assigned  value. 
If  the  assigned  value  of  the  independent  variable  be  zero, 
the  limit  is  called  the  inferior  limit ; and  if  the  value  be 
infinity,  it  is  called  the  superior  limit. 

42.  Algebraic  Illustration. — Take  the  example, 

1 


and  consider  the  series  of  values  which  y assumes  when  x 
has  assigned  to  it  different  positive  values.  When  x = 0, 
y — 1,  and  when  x has  any  positive  value,  y is  a positive 
proper  fraction ; as  x increases,  y decreases,  and  can  he  made 
smaller  than  any  assignable  fraction,  however  small,  by  giv- 
ing to  x a value  sufficiently  great.  Thus,  if  we  wish  y to  be 
less  than  roo fr-o o o>  we  make  x — 1000000,  and  get 

1 

y ~ 1 + 1000000 

which  is  less  than  100^  00  0.  If  we  wish  y to  be  less  than 
one-trillionth,  we  make  x — 1000000000000,  and  the  re- 
quired result  is  obtained.  We  see  that,  however  great  x 
may  be  taken,  y can  never  become  zero , though  it  may  be 


60 


TRIG  0 NOMETRIC  ILL  USTIiA  TION. 


made  to  differ  from  it  by  as  small  a quantity  as  we  please. 

Hence,  the  limit  of  the  function  — — — is  zero  when  x is 
infinite.  1+X 


We  are  accustomed  to  speak  of  such  expressions  thus : 
“When  x is  infinite,  y equals  zero.”  But  both  parts  of  this 
sentence  are  abbreviations:  “When  x is  infinite”  means, 
u When  x is  continually  increased  indefinitely,”  and  not, 
“ When  x is  absolute  infinity  and  “ y equals  zero  ” means 
strictly,  “y  can  be  made  to  differ  from  zero  by  as  small  a 
quantity  as  we  please.”  Under  these  circumstances,  w'e  say, 
“the  limit  of  y,  when  x increases  indefinitely,  is  zero.” 


43.  Trigonometric  Illustration — An  excellent  exam- 
ple of  a limit  is  found  in  Trigonometry.  To  find  the  values 

of  ' and  — ~r— , when  9 diminishes  indefinitely.  Here 
tan  9 9 ’ J 

we  have 

= cos  0 ; and  when  0 = 0,  cos  0 = 1. 

tan  0 

Hence,  if  0 be  diminished  indefinitely,  the  fraction 
will  approach  as  near  as  we  please  to  unity.  In  other 
words,  the  limit  of  > as  9 continually  diminishes,  is 

unity.  We  usually  express  this  by  saying,  “ The  limit  of 

sin  9 i o a * „ “sin  0 A ,j 

— - , when  0 = 0,  is  unity ; or,  : = 1,  when  0 = 0; 

tan  0 J tan  0 

that  is,  we  use  the  words  “when  0 = 0”  as  an  abbreviation 

for  “ when  0 is  continually  diminished  toward  zero.” 

Since  = 1,  when  0 = 0, 

tan  o 

= 1,  when  9 = 0. 

sin  0 


we  have  also 


TRIG  ONOMETRIC  ILL  USTRA  TION. 


61 


It  is  evident,  from  geometric  considerations,  that  if  0 be 
the  circular  measure  of  an  angle,  we  have 

tan  0 > 0 > sin  0 ; 


or, 


tan  0 0 

sin  0 sin  0 ’ 


but  in  the  limit,  i.  e.,  when  0 = 0,  we  have 
tan  0 

shT0  ~~  ’ 

and  therefore  we  have,  at  the  same  time, 

0 _ sin  0 


sin  0 


= 1, 


and 


0 


= 1, 


which  shows  that,  in  a circle,  the  limit  of  the  ratio  of  an  arc 
to  its  chord  is  unity. 

a sin  0 

In  the  expression,  - = 1,  when  0 = 0,”  it  is  evident 

that  SU^- - is  never  equal  to  1 so  long  as  0 has  a value  dif- 
ferent from  zero ; and  if  we  actually  make  0 = 0,  we  render 
the  expression  — meaningless.*  That  is,  while  ai^  ^ 

approaches  as  nearly  as  we  please  to  the  limit  unity,  it  never 
actually  attains  that  limit.  - 


If  a variable  quantity  be  supposed  to  diminish  gradually,  till  it  be 
less  than  anything  finite  which  can  be  assigned,  it  is  said  in  that  state 
to  be  indefinitely  small,  or  an  infinitesimal ; the  cipher  0 is  often  used 
as  an  abbreviation  to  denote  such  a quantity,  and  does  not  mean  abso- 
lute zero  ; neither  does  go  express  absolute  infinity. 

Rem. — The  student  may  here  read  Art.  12,  which  is  applicable  to 
this  method  as  well  as  t.o  that  of  infinitesimals,  which  it  is  not  neces 
sary  for  us  to  insert  again. 


* See  Todhunter’s  Dif.  Cal.,  p.  6. 


62 


DERIVATIVES. 


44.  Derivatives. — The  ratio  of  the  increment  of  u to 

that  of  x,  when  the  increments  are  finite,  is  denoted  by  — : 

J Ax 

the  ratio  of  the  increment  of  u to  that  of  x in  the  limit, 

dU 

i.  e.,  when  both  are  infinitely  small,  is  denoted  by  — , and 
is  called  the  derivative  * of  u with  respect  to  x. 


Thus,  let  u — f(x)  ; and  let  x take  the  increment  h 
(=  Ax),  becoming  x + h,  while  u takes  the  corresponding 
increment  A u ; then  we  have, 

u -\-  Au  — f{x  h) ; 
therefore,  by  subtraction,  we  have 

Au  -f(x  + h)  —f[x) ; 
and  dividing  by  h (=  Ax),  we  get 

Am  _ f(x  + h)  — f(x)  , . 

Ax~  h ' v ’ 

It  may  seem  superfluous  to  use  both  h and  Ax  to  denote  the  same 
thing,  but  in  finding  the  limit  of  the  second  member,  it  will  sometimes 
be  necessary  to  perform  several  transformations,  and  therefore  a sin- 
gle letter  is  more  convenient.  In  the  first  member,  we  use  Ax  on 
account  of  symmetry. 

The  limiting  value  of  the  expression  in  (1),  when  h 
is  infinitely  small,  is  called  the  derivative  of  u or 
f(x)  with  respect  to  x,  and  is  denoted  by  f (*). 


Therefore,  passing  to  the  limit,  by  making  h diminish 
indefinitely,  the  second  member  of  (1)  becomes  f '{x),  and 

du 

the  first  member  becomes,  at  the  same  time,  ; hence  we 
have 


du 

dx 


= /'(*)• 


(2) 


* Called  also  the  derived  function  and  the  differential  coefficient. 


DIFFERENTIAL  COEFFICIENT.  63 

45.  Differential  and  Differential  Coefficient. 

Let  u = then,  as  we  have  (Art.  44), 

E =/'<*>’ 

we  have  du  = df(x)  = /'  (z)  dz, 

where  efa;  and  du  are  regarded  as  being  infinitely  small,  and 
are  called  respectively  (Art.  12)  the  differential  of  x and  the 
corresponding  differential  of  u. 

f (x),  which  represents  the  ratio  of  the  differential  of  the 
function  to  that  of  the  variable,  and  called  the  derivative  of 
f(x)  (Art.  44),  is  also  called  the  differential  coefficient  of 
f(x),  because  it  is  the  coefficient  of  dx  in  the  differential 

°f  /(*)• 

Some  writers  * consider  the  symbol  ^ only  as  a whole,  and  do  not 
assign  a separate  meaning  to  du  and  dx ; others, f who  also  consider 
the  symbol  ^ only  as  a whole,  regard  it  simply  as  a convenient  nota. 

tion  to  represent  jj , and  claim  that  du  and  dx  are  each  absolutely  zero. 

46.  Differentiation  of  the  Algebraic  Sum  of  a 
Number  of  Functions. 

Let  y = au  + bv  + cw  + z + etc., 

in  which  y,  u,  v,  w,  and  z are  functions  of  x.  Suppose  that 
when  x takes  the  increment  h (—  Ax),  y,  u,  v,  w,  and  z 
take  the  increments  A y,  A u,  Aw,  A w,  A z.  Then  we  have, 

y 4.  Ay  = a (u  + A u)  + b (v  4-  Aw)  + c (tv  + Aw)  + (z  + A z)  + etc- 

Ay  = a Au  + b Aw  + c Aw  + Az  + etc. 

Dividing  by  h or  Ax , we  have 


* See  Todhunter’s  Dif.  Cal.,  p.  17  ; also  De  Morgan’s  Calculus,  p.  14,  etc. 
t See  Young’s  Dif.  Cal.,  p.  4. 


64 


DIFFERENTIATION  OF  A PRODUCT. 


Ay  A u 7 Av  Aiv  A z 

~ — a -f~  o -j~  v — — -}-  etc.. 

Ax  AX  Ax  Ax  Ax 

which  becomes  in  the  limit,  when  h is  infinitely  small 

(Art.  12), 

dy  du  , dv  dw  dz  , . . , „ 

~r  — a — ( - o + c -j h + etc.  (see  Art.  14). 

dx  dx  dx  dx  dx 


47.  Differentiation  of  the  Product  of  two  Func‘ 
tions. 


Let  y — uv,  where  u and  v are  both  functions  of  x , and 
suppose  Ay,  An,  Av  to  be  the  increments  of  y,  u,  v corre- 
sponding to  the  increment  Ax  in  x.  Then  we  have 

y + Ay  = (u  + Am)  ( v + Av) 

— uv  + u Av  + v A u + Au  Av. 


or. 


Ay  — u Av  + v Au  + Au  Av ; 

Ay  Av  Au  A u 

— = m V-  v 1 Ay. 

Ax  Ax  Ax  Ax 


Now  suppose  Ax  to  be  infinitely  small,  and 
Ay  Av  Au 

Ax ’ Ax’  Ax’ 

become  in  the  limit, 

dy  dv  du 

dx ’ dx’  dx 


Also,  since  Av  vanishes  at  the  same  time,  the  limit  of  the 
last  term  is  zero,  and  hence  in  the  limit  we  have 
dy  dv  du  /a  . , , . . 

Tx  = uTx  + vdx-  <SeeArtl6-> 

It  can  easily  be  seen  that,  although  the  last  term  vanishes,  the 
remaining  terms  may  have  any  finite  value  whatever,  since  they  con- 
tain only  the  ratios  of  vanishing  quantities  (see  Art.  9).  For  example. 

ctcc  0 ax 

— = - when  * = 0 ; but  by  canceling  x we  get  — = a.  But  the 

a:  0 ’ J ° b x 

expression  — xx,  which  equals  ^ x 0 when  x — 0,  becomes  ^ x 0 = 0 

when  x — 0. 


DIFFERENTIATION  OF  A PRODUCT. 


65 


Otherwise  thus: 

Let  f(x)  <f>  (x)  denote  the  two  functions  of  x,  and  let 
u —f{x)  <p(x). 

Change  x into  x + h,  and  let  u + Au  denote  the  new 
product;  then 

u + Au  = f(x  + h)  (p  {x  + h) 

A u _ f (x  h)  cp  (x  + h ) — f(x)  <p  (x) 

Ax  — h 

Subtract  and  add  f (x ) <p  (x  + h),  which  will  not  change 
the  value,  and  we  have 

— = /<*  + *)  -/(?) * + h)  + m ?_(•»_  + *)  - 1 W 

fi  h 

Now  in  the  limit,  when  h is  diminished  indefinitely, 

f±±R=IM  =r(x)i 

<p(x  + h)  — <p  ( x ) 


h 


— <p'  ( x ) (Art.  44) ; 


&nd 


therefore, 


(p  (x  + h)  — <P  (x) ; 
du 


fa  = /'  (*)  0 (*)  + /»  0'  (*)> 


which  agrees  with  the  preceding  result. 


48.  Differentiation  of  the  Product  of  any  Number 
of  Functions. 

Let  y = uvWy 

u,  v,  w being  all  functions  of  x. 

Assume  z = vw  ; 


then 


y — uz. 


66 


DIFFERENTIATION  OF  A FRACTION. 


and  by  Art.  47  we  have 


dy udz  zdu 

dx  ~dx  dx 


Also,  by  the  same  Article, 


dz  vdw  wdv 

dx  dx  ^ dx  * 


hence,  by  substitution,  we  have 

dy  dw  dv  du 

dr  = UV  + UW  Tx  + vw  ' (See  Art  17') 

lv*v  C I/tAj  tvd/ 


The  same  process  can  be  extended  to  any  number  ol 
functions. 


49.  Differentiation  of  a Fraction. 


Let 


Then  we  shall  have 

y + Ay 


u + Aw 
• 

v -f-  Av* 


Ay  = 


u + Au 
v + Av 


u 

V 


_ v Au  — u Av 

v1  + vAv 

Dividing  by  Ax  and  passing  to  the  limit, 

du  dv 
dy  _l  dx  U dx 
dx  ~ v 2 

(since  vdv  vanishes).  (See  Art.  18.) 

Cor. — If  u is  a constant,  we  have 

udv 

dy  _ dx 

dx  ~ v2 


DIFFERENTIATION  OF  ANY  POWER.  67 

50.  Differentiation  of  any  Power  of  a Single  Va- 
riable. 

1st.  When  n is  a positive  integer. 

Let  y — xn ; 

then  we  have  y + Ay  = (x  + h)n ; 

% l^l  

therefore,  A y = nxf~l  h H — — — - xn~2h 2 + etc. . . , h\ 

i.  • z 


Dividing  by  h or  Ax,  we  get 

^ = nx"-1  + ” ^ ~ ^ ^h...  A-1. 
Ax  2 


Passing  to  the  limit,  we  have 

(See  Art.  19,  1st.) 


dy 

~ - nx" 
dx 


(1) 


2d.  When  n is  a positive  fraction. 

m 

Let  y = u11, 

where  u is  a function  of  x ; then 

y"  — um, 

and  d ( yn ) = d ( um ) ; 

hence,  by  (1),  = tnum~ 1 

dy  m um~l  du 

dx  ~ n y"~l  dx 

= Ti  u'~l<Tx  (Art- 19’  2d)-  (2) 

3d.  When  n is  a negative  exponent,  integral  or 
fractional. 


y = 

l 

y — un> 


Let 

then 


68 


DIFFERENTIATION  OF  LOGARITHMS. 


and  by  Art.  49,  Cor.,  we  have 


dy 

dx 


nun  1 du 
idn  dx 


nu-n~l  ^ (Art.  19,  3d).  (* 


51.  Differentiation  of  log-  x. 

Let  y — log  x ; 

therefore,  y + Ay  = log  ( x + h), 

and  Ay  = log  (x  -f  h)  — log  x 

= log  Pit)  = l0«  0 + i) 

111  h2  h 3 \ 

= “h-2S+3S-etc-); 

therefore,  — = m (-  — — -f  etc.)  ; 

Ax  \x  2x2  r 

therefore,  passing  to  the  limit,  we  get 

dy  m 1 

~r~  — — or  - 
dx  x x 

(according  as  the  logarithms  are  not  or  are  taken  in  the 
Naperian  system.  See  Art.  20). 

52.  Differentiation  of  ax. 

Let  y = ax. 

Proceeding  exactly  as  in  Art.  21,  we  get 

^°g  a or  °x  a (Art.  21). 


53.  Differentiation  of  sin  x. 

Let  y = sin  x ; 

y -f  Ay  =:  sin  (x  + h) ; 

Ay  = sin  (x  + h)  — sm  x. 


therefore 

hence, 


DIFFERENTIATION  OF  A COSINE. 


69 


But  from  Trigonometry, 


hence, 


' A 'T3  o A -f  B . A — B 
sin  A-smB  = 2 cos  — - — sin  — — — 

z z 

Ay  — sin  (x  + h)  — sin  x 
( h\  . h 

h 

, sin 

Ay  I h\ 

— = cos  I x + h ) 

Ax  \ 2/ 


= 2 cos 


sm  2 ’ 


h\  Sm  2 


By  Art.  43,  when  h is  diminished  indefinitely,  the  limit  of 
= 1 ; also,  the  limit  of  cos  (x  + = cos  x. 


. h 
Sm  2 


Therefore, 


— cos  x.  (See  Art.  24.) 
ax  v 


54.  Differentiation  of  cos  x. 

Let  y — cos  x; 

therefore,  y Ay  = cos  {x  + h) ; 

hence,  Ay  = cos  {x  + h)  — cos  x 

„ . ( h\  . h 

= — 2 sm  {x  + 2)  sm  g ’ 

because  cos  A — cos  B = — 2 sin  ^ ^ 


sm 


Therefore, 


Ay 

Ax 


= -sin(*  + !) 


, h 
sin  - 


70 


COMPARISON  OF  THE  TWO  METHODS. 


Hence,  in  the  limit, 

— — sin  x.  (See  Art.  25.) 

(IX 

Of  course  this  differentiation  may  be  obtained  directly 
from  Art.  53,  in  the  same  manner  as  was  done  in  the  2d 
method  of  Art.  25. 

Since  tan  x,  cot  x,  sec  x,  and  cosec  x are  all  fractional 
forms,  we  may  find  the  derivative  of  each  of  these  functions 
by  Arts.  18  or  49,  from  those  of  sin  x and  cos  a:,  as  was  done 
in  Arts.  26,  27,  28,  and  29 ; also,  the  derivatives  of  vers  x 
and  covers  x,  as  well  as  those  of  the  circular  functions,  may 
be  found  as  in  Arts.  30,  31,  33  to  40. 

From  the  brief  discussion  that  we  have  given,  the  student 
will  be  able  to  compare  the  method  of  limits  with  the  method 
of  infinitesimals ; he  will  see  that  the  results  obtained  by 
the  two  methods  are  identically  the  same.  In  discussing  by 
the  former  method,  we  restricted  ourselves  to  the  use  of 
limiting  ratios,  which  are  the  proper  auxiliaries  in  this 
method.  It  will  be  observed  that,  in  the  former  method, 
very  small  quantities  of  higher  orders  are  retained  till  the 
end  of  the  calculation,  and  then  neglected  in  passing  to  the 
limit;  while  in  the  infinitesimal  method  such  quantities 
are  neglected  from  the  start,  from  the  knowledge  that  they 
necessarily  disappear  in  the  limit,  and  therefore  cannot 
affect  the  final  result.  As  a logical  basis  of  the  Calculus, 
the  method  of  limits  may  have  some  advantages.  In  other 
respects,  the  superiority  is  immeasurably  on  the  side  of  the 
method  of  infinitesimals. 


CHAPTER  IV. 


SUCCESSIVE  DIFFERENTIALS  AND  DERIVATIVES. 


55.  Successive  Differentials. — The  differential  ob- 
tained immediately  from  the  function  is  the  first  differential. 
Tne  differentia]  of  the  first  differential  is  the  second  differ- 
ential, represented  by  dly,  d2u,  etc.,  and  read,  “second 
differential  of  y,”  etc.  The  differential  of  the  second  dif- 
ferential is  the  third  differential,  represented  by  d3y,  cPu, 
etc.,  and  read,  “ third  differential  of  y,”  etc.  In  like  man- 
ner, we  have  the  fourth,  fifth,  etc.,  differentials.  Differen- 
tials thus  obtained  are  called  successive  differentials. 

Thus,  let  AB  be  a right  line 
whose  equation  is  y = ax  + b-, 

.*.  dy  — adx.  Now  regard  dx  as 
constant,  i.  e.,  let  x be  equicres- 
cent;*  and  let  MM',  M'M",  and 
M"M"'  represent  the  successive 
equal  increments  of  x,  or  the  dx’s, 
and  R'P',  R 'P ",  R'"P"  the  corre- 
sponding increments  of  y,  or  the 
dy’s.  'We  see  from  the  figure  that  R P'  = R"P''  = R"'P'"  ; 
therefore  the  dy’s  are  all  equal,  and  hence  the  difference 
between  any  two  consecutive  dy’s  being  0,  the  differential 

of  dy,  i.  e.,  d2y  — 0.  Also,  from  the  equation  dy adx  we 

have  iPy  — 0,  since  a and  dx  are  both  constants. 


Take  the  case  of  the  parabola  y~  = 2px  (Fig.  7),  from 
vdx 

which  we  get  dy  = ---■  Regarding  dx  as  a constant,  we 


* When  the  variable  increases  by  equal  increments,  i.  e.,  when  the  differential  it 
constant , the  variable  is  called  an  equicrescent  variable. 


72 


EXAMPLES. 


have  MM',  M'M",  M”M  " as  the  successive  equal  increments 
of  x,  or  the  dx’s ; while  we  see  from 
Fig.  7 that  RP',  R"P",  R"'P"',  or 


The  student  must  he  careful  not  to  confound  d2y  with 
dy2  or  d(y2)\  the  first  is  “second  differential  of  y,”  the 
second  is  “the  square  of  dy,”  the  third  is  the  differential 
of  y2,  which  equals  2ydy. 


1.  Find  the  successive  differentials  of  y — z5. 
Differentiating,  we  have  dy  — ox4  dx.  Differentiating 

this,  remembering  that  d of  dy  is  d2y  and  that  dx  is  con- 
stant, we  have  d2y  — 20xidx2.  In  the  same  way,  differen- 
tiating again,  we  have  d3y  — 60x2dx3.  Again,  d4y  = 120.r  dx4. 
Once  more,  d5y  = 120 dxP.  If  we  differentiate  again,  we 
have  d6y  = 0,  since  dx  is  constant. 

2.  Find  the  successive  differentials  of  y — 4.T3 — 3x2  + 2x. 


3.  Find  the  first  six  successive  differentials  of  y = sin  x. 


the  dy' s,  are  no  longer  equal,  but 
diminish  as  we  move  towards  the 
right,  and  hence  the  difference  be- 
tween any  two  consecutive  dy’ s is  a 
negative  quantity  (remembering  that 
the  difference  is  always  found  by 
taking  th e first  value  from  the  second. 
See  Art.  12).  Also,  from  the  equa- 


■x 


V 

tion  dy  = - dx  we  see  that  dy  varies  inversely  as  y. 


EXAMPLES. 


EXAMPLES. 


73 


4.  Find  the  first  six  successive  differentials  of  y — cos  x. 


)dy  — — sin  x dx  ; 
d3y  — sin  x dx3 ; 
d5y  = — sin  x dx?  ; 


d2y  — — cos  x dx2 ; 
dLy  — cos  x dxt ; 
d6y  — — cos  x dx 6. 


5.  Find  the  fourth  differential  of  y — xn. 

Ans.  d*y  = n(n  — 1)  (n  — 2)  (n  — 3)  x*~^d&. 

6.  Find  the  first  three  successive  differentials  of  y — ax. 

!dy  — ax  log  a dx  ; 
cPy  = ax  log2  a dx2 ; 
dhy  — ax  log3  a dx3. 


7.  Find  the  first  four  successive  differentials  of  y = logz. 


Ans. 


, dx 

ty  =¥; 

2 dx* 

‘py  = -¥-; 


d2y  = — 
d*y  — - 


dx2. 
x 2 ’ 
6 dx4 

xi 


8.  Find  the  first  four  successive  differentials  of  y = 2 a\fx. 


Ans.  < 


7 adx 

dV  = = > 

Vx 

7„  3 adx3 

dy  = TT ; 

4x2 


n adx 2 

cPy  = - - — r ; 

2x* 

7.  1 5adx* 

diy  = — 7-— ' 

8a:^ 


9.  Find  the  first  four  successive  differentials  of 


y = log  (1  + x)  in  the  common  system. 


Ans.  - 


cly  = 
d3y  — 


mdx 
1 + x; 
2 mdx3 


(1  + x)3’ 


d2y  = — 


mdx 2 


diy=  - 


6 mdx* 

(i  + *Y' 


10.  Find  the  fourth  differential  of  y = ex. 

Ans.  d2y  = e^dxA. 


4 


74 


SUCCESSIVE  DERIVATIVES. 


56.  Successive  Derivatives. — A first  derivative*  is 
the  ratio  of  the  differential  of  a function  to  the  differential 
of  its  variable.  For  example,  let 

y = x 6 


represent  a function  of  x.  Differentiating  and  dividing  by 
dx,  we  get 


(1) 


The  fraction  ^ is  called  the  first  derivative  of  y with 

respect  to  x,  and  represents  the  ratio  of  the  differential  of 
the  function  to  the  differential  of  the  variable,  the  value  of 
which  is  represented  by  the  second  member  of  the  equation. 


Clearing  (1)  of  fractions,  we  have 


dy  — 6 x?dx  ; 
dii 

hence,  ~ or  6x5  is  also  called  the  first  differential  coefficient 
of  y with  respect  to  x,  because  it  is  the  coefficient  of  dx. 


A second  derivative  is  the  ratio  of  the  second  differential 
of  a function  to  the  square  of  the  differential  of  the  variable. 
Thus,  differentiating  (1)  and  dividing  by  dx,  we  get  (since 
dx  is  constant,  Art.  55), 


ddy 
dx * 


= 30a4, 


(2) 


either  member  of  which  is  called  the  second  derivative  of  y 
with  respect  to  x. 


A third  derivative  is  the  ratio  of  the  third  differential  of 
a function  to  the  cube  of  the  differential  of  the  variable 
Thus,  differentiating  (2)  and  dividing  by  dx,  we  get 


* See  Arts.  44  and  45. 


DIFFERENTIAL  COEFFICIENTS. 


75 


g = 1*0*,  (3) 

either  member  of  which  is  called  the  third  derivative  of  y 
with  respect  to  x. 

In  the  same  way,  either  member  of 

§ = 360^  (4) 

dx4  v ' 

is  called  the  fourth  derivative  of  y with  respect  to  x,  and 
so  on. 

Also,  l~- , ~ , etc.,  are  called  respectively  the 
dx  dx?-  dx?  dx4  r J 

first,  second,  third,  fourth,  etc.,  differential  coefficients  of 

y with  respect  to  x,  because  they  are  the  coefficients  of  dx, 


dx2,  dx3.  dx4.  etc., 

if  (1),  (2),  (3),  (4), 

and  so 

on,  be  cleared 

of  fractions. 

In  general,  if  y 

= / ( x ),  we  have 

II 

df{x) 

dx 

= /'  (x)  (Art-  45) 

5 •••  dy 

= /'  (x)  dx. 

II 

<*/'(*) 

dx 

= f"  (*); 

•••  d?y 

= 

&•!*& 

II 

df"(x) 

dx 

=/'"(*); 

•••  d?y 

= /'"  (x)  dx3. 

II 

df"  (x) 
dx 

= /iT  0*0; 

a #y 

— f"  (x)  dx4. 

etc.  = 

etc. 

- etc. 

etc. 

- etc. 

dfy  _ 
dxf1 

df*”"  (x) 
dx 

= /(n)  (x) ; 

dry 

— f0»  (x)  dxf1. 

That  is,  the  first,  second,  third,  fourth,  etc.,  derivatives 
are  also  represented  by  /'  (x),  f (x),  f"  (x),  /iv  (x),  etc. 


76 


GEOMETRIC  REPRESENTATION. 


Strictly  speaking,  ~ or  /'  ( x ) are  symbols  representing 

the  ratio  of  an  infinitesimal  increment  of  the  function  to  the 
corresponding  infinitesimal  increment  of  the  variable,  while 
the  second  member  expresses  its  value.  For  example,  in 
the  equation  y — ax 4,  we  obtain 


g =/'(*)  = 4«A 
dv 

~ or  /'  ( x ) is  an  arbitrary  symbol,  representing  the  value 

of  the  ratio  of  the  infinitesimal  increment  of  the  function 
(ax4)  to  the  corresponding  infinitesimal  increment  of  the 
variable  ( x ),  while  4 ax?  is  the  value  itself.  It  is  usual, 
however,  to  call  either  the  derivative. 


56 a.  Geometric  Representation  of  the  First  De- 
rivative.— Let  AB  be  any  plane  curve 
whose  equation  is  y — f ( x ).  Let  P 
and  P'  be  consecutive  points,  and  PM 
and  PM'  consecutive  ordinates.  The 
part  of  the  curve  PP',  called  an  ele- 
ment* of  the  curve,  does  not  differ 
from  a right  line.  The  line  PP'  pro- 
longed is  tangent  to  the  curve  at  the 
point  P (Anal.  Geom.,  Art.  42).  Draw  PR  parallel  to  XX', 
and  we  have 

MM'  : PR  = dx,  and  RP'  = dy. 

Denote  the  angle  CTX  by  «,  and  since  CTX  = P'PR,  we 
have 

4.  dy 

tan  « = -n- 
dx 


And  since  the  tangent  has  the  same  direction  as  the  curve 


* In  this  work,  the  word  “element”  will  be  used  for  brevity  to  denote  an  “In- 
flnitesim&l  element.” 


EXAMPLES. 


77 


at  the  tangent  point  P,  « will  also  denote  the  inclination  of 
the  curve  to  the  axis  of  x. 

Hence,  the  first  derivative  of  the  ordinate  of  a curve, 
at  any  -point,  is  represented  by  the  trigonometric  tan- 
gent of  the  angle  which  the  curve  at  that  point,  or  its 
tangent,  makes  with  the  axis  of  x. 

In  expressing  the  above  differentials  and  derivatives,  we 
have  assumed  the  independent  variable  x to  be  equicrescent 
(Art.  55),  which  we  are  always  at  liberty  to  do.  This 
hypothesis  greatly  simplifies  the  expressions  for  the  second 
and  higher  derivatives  and  differentials  of  functions  of  x, 
inasmuch  as  it  is  equivalent  to  making  all  differentials  of  x 
above  the  first  vanish.  Were  we  to  find  the  second  deriva- 
tive of  y with  respect  to  x,  regarding  dx  as  variable,  we 
would  have 

cPy  _ d (dy\  _ dxd*y  — dycPx 
dx1  dx  \dx)  dx3  ’ 

o . Cp'lf 

which  is  much  less  simple  than  the  expression  j -2,  obtained 
by  supposing  dx  to  be  constant. 


EXAMPLES. 


1.  Given  y = xn,  to  find  the  first  four  successive  deriv< 
atives. 


dy 

dx 


nxn~x ; 


&y 

dx2 


n (n  — 1)  xn~2i 


dx3 


= n (n  — 1 ) [n  — 2)  xr~3 ; 


fry 
dx 4 


n (n  — 1)  {n  — 2)  (n  — 3)  z"~*. 


78 


EXAMPLES. 


If  n be  a positive  integer,  we  have 

= n(n-l)(n-2) 3.2*1. 

and  all  the  higher  derivatives  vanish. 

If  n be  a negative  integer  or  a fraction,  none  of  the  suc- 
cessive derivatives  can  vanish. 


2.  Given  y = x3  log  x ; find 

CIJ / 

= 3a;2  log  x + x2; 

CLX 


(Py 


= 6x  log  x + 3x  + 2x  = 6x  log  x + 5z; 


dx 3 


6 log  x -f  6 + 5. 


d*y  _ 6 
dx*  ~ x' 


It  can  be  easily  seen  that  in  this  case  all  the  terms  in  the 
successive  derivatives  which  do  not  contain  log  x will  dis- 
appear in  the  final  result;  thus,  the  third  derivative  of  x2  is 
zero,  and  therefore  that  term  might  have  been  neglected  ; 
and  the  same  is  true  of  5a;,  its  second  derivative  being  zero. 


1 + x , . cPy  240 

3-  prove  that  ^ 

cPu 

4.  y — eax\  prove  that  = cPe0*. 

5.  y = tan  x;  find  the  first  four  successive  derivatives. 

% = sec!*; 

^ = 2 sec2  x tan  x ; 


— 6 sec4  x — 4 sec2  x ; 
d. Xs 

= 8 tan  x sec2  x (3  sec2  x — 1). 

dx*  v 


EXAMPLES. 


79 


6. 


y = log  sin  x ; 


prove  that 


dx3 


2 cot  x cosec2  x. 


7. 


y2  = *i>x ; 


find 


cPy. 

dx? 


&y  __  d_  /p\  _ 
dx 2 ' dx  \y) 


dy^p. 

dx  y ’ 


rffJ 

*Jx 

y2 


^since 


dy 

dx  y / 


2*. 

y3’ 


d?y  _ d t p\ 

dx?  dx  \ ?/3/ 


3^§  3-*j  & 

y 6 2Z4  2/6 


8.  y = x?  \ prove  that 


jp  = sf  (1  + log  z)»  + x*-K 


9. 

aPy2 

+ J?x?  = aW 

; prove 

that 

*y. 

dx 2 

- — 

in 

X? 

• 

prove 

that 

dly 

24 

y — 

1-x9 

dx?  ' 

“ (1 

- z)5 

11. 

y2  = 

: sec  2 x ; 

prove 

that 

y + 

aj  a. 
W 

ii 

= 3^5. 

12. 

y = 

: cos  x ; 

prove 

that 

= 0. 

13. 

y = 

: x4  log  (zs)  ; 

prove 

that 

d?y 

dx? 

_ 48 
— a: 

80 


EXAMPLES. 


14.  y = x?  ; prove  that 

d?y  = 6 (dx)z  + 18 xdxd?x  -f-  3xld^xi 
when  x is  not  equicrescent. 

15.  y = / (x) ; prove  that 

d?y  = /"'  (x)  ( dxf  + 3/"  (x)  dxcPx  +/'  (x)  d*x 
when  x is  not  equicrescent. 


10,  y = ex ; prove  that 

dsy  = e?  ( dx )3  + 3eF  dx  d?x  + e?  tPx 


CHAPTER  V. 


DEVELOPMENT  OF  FUNCTIONS. 

57.  A function  is  said  to  be  developed,  -when  it  is 
transformed  into  an  equivalent  series  of  terms  following 
some  general  law. 

For  example, 

V = (a  + x)\ 

when  developed  by  the  binomial  theorem,  becomes 
y — a4  -f  4 (fix  + Qarx2  + 4 ax3  + cc4, 
which  is  a,  finite  series.  Also, 


may  be  developed  by  division  into  the  infinite  series, 

y — 1 + 2x  + 2x2  + 2X3  + etc., 

in  which  the  terms  are  arranged  according  to  the  ascending 
powers  of  x,  each  coefficient  after  the  first  term  being  2. 

One  of  the  most  useful  applications  of  the  theory  of  suc- 
cessive derivatives  is  the  means  it  gives  us  of  developing 
functions  into  series  by  methods  which  we  now  proceed  to 
explain. 


MACLAURIN’S  THEOREM. 

58.  Maclaurin’s  Theorem,  is  a theorem  for  developing 
a function  of  a single  variable  into  a series  arranged 
according  to  the  ascending  powers  of  that  variable,  with 
constant  coefficients. 


82 


MACL  AUXIN'S  THEOREM. 


Let  y =f  (x) 

be  the  function  to  be  developed  • and  assume  the  develop 
ment  of  the  form 


y = / (x)  = A + Bx  + Cx 2 + -Ox3  + j Ex*  + etc.,  (1) 

in  which  A,  B,  C,  D,  E,  etc.,  are  independent  of  x,  and 
depend  upon  the  constants  which  enter  into  the  given  func- 
tion, and  upon  the  form  of  the  function.  It  is  now  re- 
quired to  find  such  values  for  the  constants  A,  B,  C,  etc., 
as  will  cause  the  assumed  development  to  be  true  for  all 
values  of  x. 


Differentiating  (1)  and  finding  the  successive  derivatives, 
we  have. 


^ = B + 20x  + 3Dx3  + 4 Ex?  + etc,  (2) 
= 2C  + 2 • Wx  + 3 • 4 Ex?  + etc.,  (3) 


= 2 • 3D  + 2 ■ 3 • 4Ex  + etc.,  (4) 

ctx5 

S - 2 • 3 • ±E  + etc.,  (5) 


Now,  as  A,  B , C,  etc.,  are  independent  of  x,  if  we  can 
find  what  they  should  be  for  any  one  value  of  x,  we  shall 
have  their  values  for  all  values  of  x.  Hence,  making  x — 0 
in  (1),  (2),  (3),  etc.,  and  representing  what  y becomes  on 

this  hypothesis  by  (y)  ; what  ^ becomes  by  ; what 


(Py 

dxP 


becomes 


dx 

by  5 and  so  on  ; we  have, 


(y)  = A; 


A = (y)- 

s = © 


MACLAURIN’S  THEOREM. 


83 


\dx2) 

II 

.•  C = 

(ac3) 

CO 

II 

.-.  D = 

(&y\ 

\dxi/ 

= 2 • 3 • 4A; 

.-.  E = 

3 -4 


(£py\  i 

\dxV  1-2-3' 

ldly\  1 
\dxdj  TTJT 

Substituting  these  values  in  (1),  we  have, 

r = ,w  = <,)  + (g  H + ^ + 

+ (S)  nsvri  + (•) 


which  is  the  theorem  required. 


Hence,  by  Maclaurin’s  Theorem,  we  may  develop  a func- 
tion of  a single  variable,  as  y — f{x),  into  a series  of  terms, 
the  first  of  which  is  the  value  of  the  function  when  x = 0 ; 
the  second  is  the  value  of  the  first  derivative  of  the  function 
when  x = 0 into  x ; the  third  is  the  value  of  the  second 

cc^ 

derivative  when  x = 0 into  , etc.;  the  (n-\-l)th  term  is 

/V 


a?1 


1-2-3  . . n \dxn, 


We  may  also  use  the  following  notation  for  the  function 
and  its  successive  derivatives  : f (x),  f (x),  f"  ( x ),  f"  (x), 
f"(x),  etc.,  as  given  in  Art.  56,  and  write  the  above 
theorem, 

v=m=  m + r w f + r m ~ + r <o> 

+ /"  ((>)  iraTTTi  + etc->  f‘) 


in  which  /( 0),  /'  (0),  f"  (0),  f"  (0),  etc.,  represent  the 
values  which  f(x ) and  its  successive  derivatives  assume 


84 


MACLAURIN’S  THEOREM. 


when  x = 0.  We  shall  use  this  notation  instead  of 
(te , ~ , etc.,  for  the  sake  of  brevity. 

This  theorem,  which  is  usually  called  Maclaurin’s  Theorem,  was 
previously  given  by  Stirling  in  1717  ; but  appearing  first  in  a work  on 
Fluxions  by  Maclaurin  in  1742,  it  has  usually  been  attributed  to  him, 
and  has  gone  by  his  name.  Maclaurin,  however,  laid  no  claim  to  it, 
for  after  proving  it  in  his  book,  he  adds,  “ this  theorem  was  given  by 
Dr.  Taylor.”  See  Maclaurin’s  Fluxions,  Vol.  2,  Art.  751. 

To  Develop  y = (a  + a?)6 

f(x)  = (a  + z)6; 

/( 0)  = «6- 
/'  ( x ) = 6 (a  + xf ; 

/'( 0)  = 6a*. 
f"  (x)  = 5 • 6 (a  -f  x)*; 
f"  (0)  r=  5 • 6a4. 
f"(x)  = 4-  5.  6 (a  + xf; 

/"'  (0)  = 4 • 5 • 6as. 
f(t)  = 3-4-5.6(a  + z)3; 

/,T  (0)  = 3-4.5-  6a2. 

/'  (x)  = 2 • 3 - 4 - 5 - 6 (a  + x) ; 

/» (0)  = 2 • 3 ■ 4 • 5 ■ 6a  ; 

f*(x)  = 1 • 2-  3-  4-  5 - 6; 

f*  (0)  = 1 - 2 • 3 - 4 ■ 5 - 6 ; 

Substituting  in  (7),  we  have, 

j = (a  + x)6  — a6  + 6 a5x  + 5 • 6a4- — ^ + 4 • 5 • 60^  .2 .3 

cfix*  2- 3- 4- 5- 6 a-T5  1- 2- 3- 4- 5- 6z« 

f 3 • 4 - 5 • ° x . 2 . 3 . 4 + l-2-3-4-5  + l-  2-  3-  4-5-6 

= a6  + 6a5z  + 15a4z2  4-  20asz3  + 15a V + Gax5  +X6, 


Here 

lence. 


THE  BINOMIAL  THEOREM. 


85 


which  is  the  same  result  we  would  obtain  by  the  binomial 
theorem. 


THE  BINOMIAL  THEOREM. 

59.  To  Develop  y = (a  + x)\ 

Here  / ( x ) = (a  + x)n  ; 

hence,  /( 0)  = a". 

f (x)  — n (a  + a:)71-1 ; 

f (0)  = nan~K 

f"(x)  = n (n  — 1)  {a  -f  a;)"-3. 
f"  (0)  — n (n  — 1)  an~2. 
f"  ( x ) = n {n  — 1)  (n  — 2)  (a  + x)n~3 ; 

f"  (0)  = n (n  — 1)  ( n — 2)  a'1-3. 

/iT( x ) = n (n  — 1)  ( n — 2)  ( n — 3)  (a  + a:)"-4; 

f 'n  (0)  = n (n  — 1)  (n  — 2)  (n  — 3)  a"-4,  etc. 


Substituting  in  (7),  Art.  58,  we  have, 

, . , n (n  — 1)  an~2xl 

y = (a  + x)n  = an  + 7ian~xx  -\ — - — 


n (n  — 1)  (n  — 2)  an~3x3 
lT2"^3  _ 

n (n  — 1)  (n  — 2)  (n  — 3)  fln_4a^ 

1 • 2 • 3 • 4 


+ etc. 


Thus  the  truth  of  the  binomial  theorem  is  established, 
applicable  to  all  values  of  the  exponent,  whether  positive 
or  negative,  integral  or  fractional,  real  or  imaginary. 


60.  1.  To  Develop  y = sin  x. 

Here  f(x)  = sin  x ; hence,  /( 0)  = 0. 

“ /'  (x)  = cos  X ; “ /'  (0)  = 1. 

**  f"  ( x ) = — sin  x ; “ f"  (0)  = 0. 


86 


THE  LOGARITHMIC  SERIES. 


Here  /'"  ( x ) = — cos  x ; 
“ /iT  (x)  — sin  x ; 

“ P ( x ) — cos  x ; 

Etc.,  etc. 


hence,  / " (0)  = — "L 
“ f (0)  = 0. 

“ P(  0)=1. 

Etc.,  etc. 


Hence, 


x? 

y = sin  x = x — lm2m2  + 


X5 

1 • 2 • 3 • 4-  5 


x~ 

1- 2-3- 4- 5-6-7 


+ etc. 


2.  To  Develop  y = cos  x. 

A X2 

Am.  y = cos  z = 1 - j-j  + 

X?  X? 

~ 1 . 2 • 3 • 4 • 5 • 6 + 1- 2-3-4- 5- 6- 7-8  ~ etC< 

The  student  will  observe  that  "by  taking  the  first  derivative  of  th« 
series  in  (1),  we  obtain  the  series  in  (2),  which  is  clearly  as  it  should 
be,  since  the  first  derivative  of  sin  x is  equal  to  cos  x. 

Since  sin  (—  x)  — — sin  x,  from  Trigonometry  we  might  have 
inferred  at  once  that  the  development  of  sin  x in  terms  of  x could  con- 
tain only  odd  powers  of  x.  Similarly,  as  cos  (—  X)  — cos  x.  the 
development  of  cos  x can  contain  only  even  powers. 

By  means  of  the  two  formulae  in  this  Article  we  may 
compute  the  natural  sine  and  cosine  of  any  arc.  For  exam- 
ple, to  compute  the  “natural  sine  of  20°,  we  have  x — arc  of 

20°  = - = .3490652,  which  substituted  in  the  formulae, 
9 

gives  sin  20°  .342020  and  cos  20°  = .939693. 


THE  LOGARITHMIC  SERIES. 

61.  To  Develop  y — log  (1  + x)  in  the  system  in 
which  the  modulus  is  m. 

Here  f(x)  = log  (1  + x);  hence,  /( 0)  = 0. 

" /'(*)  = = 


TEE  LOGARITHMIC  SERIES. 


87 


Ifl 

Her ef"(x)  = - hence,/'  (0)  = - m. 


1.0.  Qra 

“ /iT  ; “ /iv  (°)  = -'• 2- 3m- 


Etc. 


Etc. 


Substituting  in  (7),  Art.  58,  we  have, 

y = log  (1  -\-x)  — m (x— + lad  + ^r5— etc.),  (1) 

which  is  called  the  logarithmic  series. 

Since  in  the  Naperian  system  m — 1 (see  Art.  20,  Cor.), 
we  have. 


y = \og{l  + x)  = x — 2+3  — 4+5  — etc‘  <-2) 


which  is  called  the  Naperian  logarithmic  series. 

This  formula  might  be  used  to  compute  Naperian  loga- 
rithms, of  very  small  fractions ; but  in  other  cases  it  is 
useless,  as  the  series  in  the  second  number  is  divergent  for 
values  of  x > 1.  We  therefore  proceed  to  find  a formula 
in  which  the  series  is  convergent  for  all  values  of  x ; i.  e.,  in 
which  the  terms  will  grow  smaller  as  we  extend  the  series. 

Substituting  — x for  x in  (2),  we  have, 


log  (1  + x)  — log  (1  — x)  = 2x  + — g-  + -y  + -y  + etc- 


X2  X3  X4  X5 


Subtracting  (3)  from  (2),  we  have, 


2x®  2z5  2a:7 


88 


CALCULATION  OF  LOGARITHMS. 


1 + X _ Z + 1 
1 — X Z 


Substituting  in  (4),  we  have, 


+ 3 (2 2 + l)3  + 


1 


l)3  + 5 (2z  + l)5 


1 


or 


log  (z  + 1)  = log  z + 2 [2z  ^ 1 


1 + 3 (2z  + l)3 


1 


+ 5 (2z  + l)5  + 


This  series  converges  for  all  positive  values  of  z,  and  more 
rapidly  as  z increases.  By  means  of  it  the  Naperian  loga- 
rithm of  any  number  may  be  computed  when  the  logarithm 
of  the  preceding  number  is  known.  It  is  only  necessary  to 
compute  the  logarithms  of  prime  numbers  from  the  series, 
since  the  logarithm  of  any  other  number  may  be  obtained 
by  adding  the  logarithms  of  its  factors.  The  logarithm  of  1 
is  0.  Making  z = 1,  2,  4,  G,  etc.,  successively  in  (5),  we 
obtain  the  following 

Naperian  or  Hyperbolic  Logarithms. 


\ .00082305  f 

log  2 = 2 / . 00006532  ) = 2(0.34657359)  = 0.6931471& 
) .00000565  ( 


or,  since 


log  1 = 0. 


.33333333 

.01234568 


.00000051 

.00000005 


CALCULATION  OF  LOGARITHMS. 


89 


log  3 = log  2 + 2 (-  + — 3 + — 5 + — 7 + ^59  + etc.) 

= 1.09861228. 

log  4 = 2 log  2 : - 1.38629436. 

log  5 = log  4 +•  2 (-  + — 3 + g7g-5  + 7^797  + 9799  + etc-) 

= 1.60943790. 

log  6 = log  3 + log  2 = 1.79175946. 

iog  7 = log  6 + 2 (—  + — p + gT^gs  + ^7jg7  + etc.) 

= 1.94590996. 

log  8 = 3 log  2 = 2.07944154. 

log  9 = 2 log  3 = 2.19722456. 

log  10  = log  5 + log  2 = 2.30258509. 

In  this  manner,  the  Naperian  logarithms  of  all  numbers 
may  be  computed.  Where  the  numbers  are  large,  their 
logarithms  are  computed  more  easily  than  in  the  case  of 
small  numbers.  Thus,  in  computing  the  logarithm  of  101, 
the  first  term  of  the  series  gives  the  result  true  to  seven 
places  of  decimals. 

Cor.  1. — From  (1)  we  see  that,  the  logarithms  of  the 
same  number  in  different  systems  are  to  each  other 
as  the  moduli  of  those  systems;  and  also,  that  the 
logarithm  of  a number  in  any  system  is  equal  to  the 
Naperian  logarithm  of  the  same  number  into  the 
modulus  of  the  given  system. 

Cor.  2. — Dividing  (1)  by  (2),  we  have 

Common  log  (1  + x)  _ ^ 

Naperian  log  (1  + x)  ' ' 

Hence,  the  modulus  of  the  common  system  is  equal 
to  the  common  logarithm  of  any  number  divided  by 
the  Naperian  logarithm  of  the  same  number. 


90 


EXPONENTIAL  SERIES. 


Substituting  in  (6)  the  Naperian  logarithm  of  10  com- 
puted above,  and  the  common  logarithm  of  10,  which  is  1, 
we  have 

m = o qaoLkao  = .1342944819032518276511289  . . . 

2.30258o09 

which  is  the  modulus  of  the  common  system.  (See  Serret’s 
Calcul  Differentiel  et  Integral,  p.  169.) 

Hence,  the  common  logarithm  of  any  number  is 
equal  to  the  Naperian  logarithm  of  the  same  number 
into  the  modulus  of  the  common  system, 

Cor.  3. — Representing  the  Naperian  base  by  e (Art.  21, 
Cor.  2),  we  have,  from  Cor.  1 of  the  present  Article, 

com.  log  e : Nap.  log  e ( = 1)  ::  .43429448  : 1 ; 
therefore,  com.  log  e = .43429448 ; 
and  hence,  from  the  table  of  common  logarithms,  we  have 
e = 2.718281  + . 


EXPONENTIAL  SERIES. 


62. 

To  Develop  y = 

ax. 

Here 

II 

K 

hence, 

/( 0)  = 

1. 

66 

f (x)  = ax  log  o 

: ; “ 

/'( 0)  = 

log  a. 

u 

f"  (*)  = ax  (log 

a)2;  “ 

/"( 0)  = 

(log  af 

66 

. /"'  (x)  = ax  (log  ( 

*)3;  “ 

/"'( 0)  = 

(log  a)3. 

■d  the  development  is 

9 : 

X 

— ax  = 1 + log  a - 

+ log2  a ^ 

+ log3  a 1 

Xs 

• 2-3 

a* 

1-2-3-4 


+ etc. 


(1) 


4-  log4  a 


EXPONENTIAL  SERIES. 


91 


Cor. — If  a = e,  the  Naperian  base,  the  development 
becomes 


Putting  x — 1,  we  obtain  the  following  series,  which  en- 
ables us  to  compute  the  value  of  the  quantity  e to  any 
required  degree  of  accuracy : 


= 2.718281828 + . 

63.  To  Develop  y = tan-1  x. 

In  the  applications  of  Maclaurin’s  Theorem,  the  labor  in 
finding  the  successive  derivatives  is  often  very  great.  This 
labor  may  sometimes  be  avoided  by  developing  the  first 
derivative  by  some  of  the  algebraic  processes,  as  follows : 

Here  f{x)  = tan ~x  x\ 

hence,  f (0)  = 0. 


= (lay  division)  1 — x2  + x4  — + 

/'(0)  = 1- 

f"  (. x ) = — 2x  4.r3  — Gx5  8x 7 — lOx9  -f  etc. ; 

/"( 0)  = 0. 

f"  (x)  — — 2 + 3 • 4x2  — 5 • Gx4  + 7 • 8.T6  — etc. ; 
/"'  (0)  = - 2. 

,f"(x)  = 2-3-4z  — 4-5- 6a:3  + etc.; 

/w(0)  - o- 


+ • ' • • 1.2-37'..  « + etC-  (2) 


, 111  1 
y_e_2  + - + ~-  + 2T77  + 2^ 


2-3- 4- 5 


1 


92 


FAILURE  OF  MACLAURIN’S  THEOREM, 


f'(x)  = 2-3-4  — 3 • 4 • 5 • 6a:2  + etc.; 
f (0)  = 2-3-4. 

fiv'  {x)  = — 2- 3 -4- 5- 62  + etc. ; 

r (o)  = o. 

/™  (x)  — — 2- 3 -4- 5- 6 + etc. ; 

/™  (0)  — — 2- 3 -4- 5 -6. 

Substituting  in  (7)  of  Art.  58,  we  get 

/y*3  /y<5 

y — tan-1  x — x — g+5  - ^ + etc. 


64.  It  sometimes  happens  in  the  application  of  Maclau- 
rin’s  Theorem  that  the  function  or  some  of  its  derivatives 
become  infinite  when  x — 0.  Such  functions  cannot  be 
developed  by  Maclaurin’s  Theorem,  since,  in  such  cases, 
some  of  the  terms  of  the  series  would  be  infinite,  while  the 
function  itself  would  be  finite. 

For  example,  take  the  function  y = log  x.  Here  we 
have 

f{x)  —\ogx-,  hence,  /( 0)  = — go. 

f\x)=\\  “ /'( 0)  = c°. 


f"(x)  = 


“ /"( 0)  = -». 


etc. 


etc. 


Substituting  in  Maclaurin’s  Theorem,  we  have 

X ccP 

y = log  x — — oo  + co  - — oo—  -f-  etc. 

Here  we  have  the  absurd  result  that  log  x = oo  for  all 
ralues  of  x.  Hence,  y — log  x cannot  be  developed  by 
Maclaurin’s  Theorem. 

Similarly  y — cot  x gives,  when  substituted  in  Maclau- 
■*Vrds  Theorem, 


taylor’s  theorem. 


93 


, x 

y — cot  x = oo  — oo  - -j-  etc. ; 

that  is,  cot  x = oo  for  all  values  of  x , which  is  an  absurd 
result.  Hence,  cot  x cannot  be  developed  by  Maclaurin’s 
Theorem. 

Also,  y — x?  becomes,  by  Maclaurin’s  Theorem, 
y — x*  — 0 + cox  + etc. ; 

that  is,  x%  = oo  for  all  values  of  x,  which  is  an  absurd 
result. 

Whether  the  failure  of  Maclaurin’s  Theorem  to  develop  correctly  is 
due  to  the  fact  that  the  particular  function  is  incapable  of  any  devel- 
opment, or  whether  it  is  simply  because  it  will  not  develop  in  the 
particular  form  assumed  in  this  formula,  the  limits  of  this  book  will 
not  allow  us  to  enquire. 


TAYLOR’S  THEOREM. 

65.  Taylor’s  Theorem  is  a theorem  for  developing  a 
function  of  the  sum  of  two  variables  into  a series  arranged 
according  to  the  ascending  powers  of  one  of  the  variables, 
with  coefficients  that  are  functions  of  the  other  variable  and 
of  the  constants. 

Lemma. — We  have  first  to  prove  the  following  lemma: 
If  we  have  a function  of  the  sum  of  two  variables  x and  y, 
the  derivative  will  be  the  same,  whether  we  suppose  x to 
vary  and  y to  remain  constant,  or  y to  vary  and  x to  remain 
constant.  For  example,  let 

u = (x  + y)n.  (1) 

Differentiating  (1),  supposing  x to  vary  and  y to  remain 
constant,  we  have 


— = n(x  + yY~' 


(2) 


u 


TAYLOR’S  THEOREM. 


Differentiating  (1),  supposing  y to  vary  and  x to  remain 
constant,  we  have 

du  . . . . . 

y-y  = n(x  + yf~l;  (3) 

from  which  we  see  that  the  derivative  is  the  same  in  both 
(2)  and  (3). 


In  general , suppose  we  have  any  function  of  x + y,  as 

«=/(*  + y)-  (4> 

Let  z — x + y ; (5) 

•••  u=f(z).  (6) 

Differentiating  (5),  supposing  x variable  and  y constant, 
and  also  supposing  y variable  and  x constant,  we  get 


dz  

dx  ~ ’ 


and  = X. 

dy 


Differentiating  (6),  we  have 

= =/■(,).  (See  Art.  45.) 


dz 


da  = f (z)  dz. 


du 


(si 


dz 

.since  ~r~ 
\ dx 


= 4 


And  similarly, 

S =/<=/(*)  (since  | =!). 


dy 

du 

dx 


du 

dy' 


That  is,  the  derivative  of  u with  respect  to  x,  y being 
constant,  is  equal  to  the  derivative  of  u with  respect 
to  y,  x being  constant. 


METHOD  OF  TAYLOR'S  THEOREM. 


95 


66.  To  prove  Taylor’s  Theorem. 


Let  u'  =f(x  + y)  be  the  function  to  be  developed,  and 
assume  the  development  of  the  form 


u’  = f{x  + y) 

— A + By  + Cif  + Dif  + Ey 4 + etc.,  (1) 


in  which  A,  B,  C,  etc.,  are  independent  of  y,  but  are  func- 
tions of  x and  of  the  constants.  It  is  now  required  to  find 
such  values  for  A,  B,  C,  etc.,  as  will  make  the  assumed 
development  true  for  all  values  of  x and  y. 

Finding  the  derivative  of  u,  regarding  x as  constant  and 
y variable,  we  have 

rj 7/ 1 

At  = B + 2Cy  + 3 Dffi  + ±Etf  + etc.  (2) 

Again,  finding  the  derivative  of  u’,  regarding  x as  varia- 
ble and  y constant,  we  have 


dn' 

dx 


dB 

dx 


dC 


dB 


^ + ^2  + ^3  + etc- 


(3) 


By  Art.  65,  we  have  = ~ ; therefore, 

J dy  dx 

■n  , nr y ono  a r,  o , dA  dB  dC  „ dD  „ 

B+ZCy  + iDf  + Wf+z  te.  = - + 


+ etc. 


(4) 


Since  (1)  is  true  for  every  value  of  y,  it  is  true  when 
y — 0.  Making  y — 0 in  (1),  and  representing  what  u' 
becomes  on  this  hypothesis  by  u,  we  have 


« = f(x)  = A- 


(5) 


Sinoe  (4)  is  true  for  every  value  of  y,  it  follows  from  the 
principle  of  indeterminate  coefficients  (Algebra)  that  the 
coefficients  of  the  like  powers  of  y in  the  two  members 
must  be  equal.  Therefore, 


96 


TAYLOR’S  THEOREM. 


B = 
2C  = 


4 E — 


have 


dA 
dx  ’ 

B = %’Smm  <5>; 

dB 

• 

„ 1 cPu 

" ~ D2  ' 

dx  ’ 

dC 

jj  1 d?u 

~ 1-2-3  ' dx?  ’ 

dx  ’ 

dB 

p 1 

dx  ’ 

' 1-2-3-4  dx? 

ng  these  values  of 

A,  B,  C,  D,  etc.,  in  (1), 

u>  _ f(x  , v)  _ , du  y d*u  y*  d?u  y* 

U ~J{X  + y)  ~U  + dxl  + dx>l-2  + d&l^: 


+ 


dhi  yi 


dx?  1-2- 3-4 


+ etc. 


(6) 


Or.  using  the  other  notation  (Art.  56),  we  have 
u'  = f(x  + y)  =f(x ) +/'(z)|  +f"{x)  +f"'{x)^t- _ 

+/,v  (a;)  r&4 + etc-j  (7) 


which  is  Taylor’s  Theorem.  It  is  so  called  from  its  discov- 
erer, Dr.  Brook  Taylor,  and  was  first  published  by  him  in 
1715,  in  his  Method  of  Increments. 


Hence,  by  Taylor’s  Theorem,  we  may  develop  a function 
of  the  sum  of  two  variables,  as  u =f(x  -f-  y),  into  a series 
of  terms,  the  first  of  which  is  the  value  of  the  function 
when  y = 0 ; the  second  is  the  value  of  the  first  derivative 
of  the  function  when  y = 0,  into  y ; the  third  is  the  value 

of  the  second  derivative  when  y — 0,  into  etc. 


The  development  of  f{x  — y)  is  obtained  from  (6)  or  (7). 
hy  changing  4-  y into  — y ; thus. 


BINOMIAL  THEOREM. 


97 


f(x  - y ) 


du  y cPu  y2  d3u  y3 

dx  1 dx 2 1 • 2 dx?  1 • 2 • 3 


dhi  y4 


— etc. 


or,  fix  -y)=  f(x)  -f  (x)  | +/"  (;t)  ^ -f  ix)  ^ 

+/iT('r)r&i_etc- 

Cor. — If  we  make  x = 0 in  (7),  we  have 

«'  = /to)  = /(0)  + f (o)  f + /"  (0)  ^ + /'"  (o)  j-A 

+ -/'"<0,lT&‘i  + etc-’ 

which  is  Maclaurin’s  Theorem.  See  (7)  of  Art.  58. 


THE  BINOMIAL  THEOREM. 


67.  To  Develop  u'  = (ac  + ?/)". 

Making  y = 0,  and  taking  the  successive  derivatives,  we 
have 

f(x)  = a;n, 

/'  (a:)  = nxn~\ 

f"  (x)  = n (n  — 1)  a:"-2, 

/'"  (a-)  = n (7i  — 1)  (%  — 2)  a;”-3, 

/iT  (a:)  = » (w  — 1)  (w  — 2)  (w  — 3)  af-4, 
etc.  etc. 


Substituting  these  values  in  (7),  Art.  66,  we  have 

, , . nx,‘~1  y n(n  — 1)  xn~ 2 y 2 

« = (»  + */)n  = a"  + — ^ + -| jTa “ 

n{n  — 1)  (n  — 2 )xn~3y3  , 

+ 1-2.3  -+etc., 

which  is  the  Binomial  Theorem  (see  Art.  59). 

5 


98 


APPLICATIONS  OF  TAYLOR'S  THEOREM. 


68.  To  Develop  ur  — sin  {pc  + y). 

Here  / (x)  = sin  x,  f ( x ) = cos  x, 

f"  (x)  — — sin  x,  /"'  ( x ) — — cos  x,  etc. 

Hence, 

v!  — sin  {x  + y) 

A y2  w4  yG  \ 

= sm n1  - + riz- 4 - TYs-  are  + etc'; 


+ cos  x 


y y 

H2T3  + 1-2-3-4-5 


f 


(y 

\L  1-2-3  1 1-2-3-4-5  1-2-3-4-5-6-7 

= sin  a:  cos  y + cos  a;  sin?/.  (See  Art.  60.) 


+ etc. 


I 


THE  LOGARITHMIC  SERIES. 

69.  To  Develop  ur  = log-  (pc  + y). 


Here  f ( x ) = log  x, 

ro)  = \. 


/"(*)  = 


Hence, 


v!  — log  (x  + y) 
= log  X + - - 


/'"(*)  = 1* 
etc. 


1 -w2  1 w3  1 ?/4 

k ~ o 4 o i — j b 4-  eto 

2 a:2  3 a4  4 a,4 


Cor. — If  x—1,  this  series  becomes 

log  (1  4-  y)  = | — | + | — j 4-  etc., 
which  is  the  same  as  Art.  61. 


EXPONENTIAL  SERIES. 

70.  To  Develop  u’  — cix+v. 

Here  f(x)  = ax,  f"  (x)  = ax  log2  a, 

f'(x)  — ax  log  a,  (x)  — ax  log3  a,  etc. 


FAILURE  OF  TAYLOR’S  THEOREM. 


99 


Hence, 

u'  — ax+y 

= ax(l  + lo  ga-y  + log  2a~^  + + etc-)- 

Coe. — If  x = 0,  this  series  becomes 
ay  = 1 + log  a-y  + log2a^  + loS3af^3  + etc-> 
which  is  the  same  as  Art.  62. 


71.  Though  Taylor’s  Theorem  in  general  gives  the  cor- 
rect development  of  every  function  of  the  sum  of  two 
variables,  yet  it  sometimes  happens  that,  for  particular 
values  of  one  of  the  variables,  the  function  or  some  of  its 
derivatives  become  infiyiite  ; for  these  particular  values,  the 
theorem  fails  to  give  a correct  development. 


For  example,  take  the  function  v!  — V a + x + y. 
Here,  f(x)  — Va  + x, 

f (*) 


/•"(*)  = - 
/'"  (*)  = 


2 V a + x 

1 


4 (a  + x)? 
3 


_ . ,s  ) 

8 (a  + 


etc. 


Substituting  in  (7)  of  Art.  66,  we  have 


v!  = Va  + » + y 


= V a + x + 


+ 


yS 


2 V a -\- x 8 (a  + xp  16(u  + a;)v 


•etc. 


Now  when  x has  the  particular  value  — a,  this  equation 
becomes 

u'  = vV  = 0 4-  oo  — co  + oo  — etc. ; 


100 


EXAMPLES. 


that  is,  when  x = — a,  Vy  = ° o . But  y is  independent  oi 
x,  and  may  have  any  value  whatever,  irrespective  of  the 
value  of  x,  and  hence  the  conclusion  that  when  x = — a, 
V y = oo,  cannot  be  true.  For  every  other  value  of  x, 
however,  all  the  terms  in  the  series  will  be  finite,  and  the 
development  true. 

Similarly,  u'  = a + V a — x + y gives,  when  substituted 
in  Taylor’s  Theorem, 

ul  = a + V a — x + y 

= a + V a — x - -f-  etc., 

2V  a — x 

which,  when  x = a,  becomes 

vl  = a + \/ y = a — oo  + etc. ; 

and  hence  the  development  fails  for  the  particular  value, 
x = a. 

It  will  be  seen  that  when  Taylor’s  Theorem  fails  to  give 
the  true  development  of  a function,  the  failure  is  only  for 
particular  values  of  the  variable,  all  other  values  of  both 
variables  giving  a true  development ; but  when  Maclaurin’s 
Theorem  fails  to  develop  a function  for  one  value  of  the 
variable,  it  fails  for  every  other  value. 

Many  other  formulae,  still  more  comprehensive  than  these, 
have  been  derived,  for  the  development  of  functions;  but  a 
discussion  of  them  would  be  out  of  place  in  this  work. 


EXAMPLES. 

1.  Develop  y = Vl  + x2. 

Put  x 2 — z,  and  develop ; then  replace  z by  its  value. 
Ans.  y = Vl  + %% 


, o?  xi  x?  5z® 

“ + 2 ~ 8 + 16  _ 128  + etC* 


* 


EXAMPLES. 


101 


2.  y = 

y - 

s.  y = 
y = 

4.  y = 

y = 

5.  y = 

y = 

6.  y = 

y = 

7.  y = 
y = 

8.  y = 
y = 

9-  y = 

Put 

y = 


i 


1 — £ 


l + a^a^  + a^-fa^-f-  etc. 


(«  + z)-s. 

(a  + a;)-3  = a-3  — 3a_4a;  + 6a_5a“2  — lOa-^3 

f-  etc. 

gSin 

. z2  ar4  a^ 

«8,DI  + 


2 2*4  3-5  2-4-5. 1 

+ etc. 


xe?. 

. z3  a4 

a^*  = a;  + a:2  + - + — ^ + etc- 


V^a;  — 1. 

V2a;  — 1 - V—  l(l—  z — ^ ~ — etc.^. 

(a2  + x2)%. 

(a2  + x2)i  — a ^ + §a*x2  + fyar^x*  — -fca  §xP 

+ etc. 

_1 

a4  + x* 

1 1 a4  5a?  5- 9a;12 

(a4  + :c  ) — ^ — 4^5  + 478fli  — 4-  8-  12a13 


+ 


5 -9-  13a:16  ^ 

4- 8- 12- 16a17  etC> 


(a5  + a4x  — x5)^. 

a4x  — x5  = z,  as  in  Ex.  1. 

x 4 x2  4-9  x3 
a + 5~5hT2+5 W'  1-2-3 

4-9-14  x4 
5%3  " 1-2-3-4 


102 


EXAMPLES. 

10.  u — (x  + y )?. 

u = & + + -h  x~*ys  — eta 

11.  u = cos  (x  + y ).  (See  Art.  68.) 

u = cos  {x  + y)  = cos  a;  cos  y — sin  a:  sin  y. 

12.  y — tan  x. 

y — tan  x = x + ~+  — + etc. 

13.  y — sec  a:. 

a^  5a^  61a^ 

3,  = see  =;  = 1 + ^ +_+_  + ete. 

14.  y = log  (1  + sin  x). 

/£ 2 ^3  ^4 

Jf  = log  (1  + sin  x)  = x — + g — + etc. 


CHAPTER  VI. 


EVALUATION  OF  INDETERMINATE  FORMS. 


72.  Indeterminate  Forms. — When  an  algebraic  ex- 
pression is  in  the  form  of  a fraction,  each  of  whose  terms  is 
variable,  it  sometimes  happens  that,  for  a particular  value 
of  the  independent  variable,  the  expression  becomes  inde- 
terminate; thus,  if  a certain  value  a when  substituted  for 


x makes  both  terms  of  the  fraction 
0 


/(*) 


vanish,  then  it 


reduces  to  the  form 
nate. 


O’ 


<p  {x) 

and  its  value  is  said  to  be  indetermi- 


Similarly,  the  fraction  becomes  indeterminate  if  its  terms 
both  become  infinite  for  a particular  value  of  x;  also  the 
forms  go  x 0 and  go-go  , as  well  as  certain  others  whose 
logarithms  assume  the  form  oo  x 0,  are  indeterminate  forms. 
It  is  the  object  of  this  chapter  to  show  how  the  true  value 
of  such  expressions  is  to  be  found.  By  its  true  value  is 
meant  the  limiting  value  which  the  fraction  assumes  when 
x differs  by  an  infinitesimal  from  the  particular  value  which 
makes  the  expression  indeterminate.  It  is  evident  (Arts.  9, 
43)  that  though  the  terms  of  the  fraction  may  be  infinitesi- 
mal, the  ratio  of  the  terms  may  have  any  value  whatever. 

In  many  cases,  the  true  values  of  indeterminate  forms 
can  be  best  found  by  ordinary  algebraic  and  trigonometric 
processes. 

For  example,  suppose  we  have  to  evaluate  ^ when 

x =1.  This  fraction  assumes  the  form  jj  when  x = 1 ; 
but  if  we  divide  the  numerator  and  denominator  by  a:  — 1 


104 


EXAMPLES. 


before  making  x — 1,  the  fraction  becomes  — x + l — ’ 

and  now  if  we  make  x = 1,  the  fraction  becomes 

1 + 1 + 1 _ 3 
1 + 1 2’ 

which  is  its  true  value  when  x — 1. 

73.  Hence  the  first  step  towards  the  evaluation  of  such 
expressions  is  to  detect,  if  possible,  the  factors  common  to 
both  terms  of  the  fraction,  and  to  divide  them  out;  and 
then  to  evaluate  the  resulting  fraction  by  giving  to  the 
variable  the  assigned  value. 


EXAMPLES. 


1.  Evaluate 


x3  — 1 

x3  — 2x2  + 2x  — 1 ’ 


when  x = 1. 


This  fraction  may  be  written 


(x  — 1)  (x2  + x + 1)  _ x2  + x + 1 
[x  — 1)  (x2  — x + 1)  ~ x2  — x + 1 


- - 3,  when  x = 1. 


x 0 

2.  The  fraction  . when  x = 0. 

V a + x — v a — x u 

k 

To  find  its  true  value,  multiply  both  terms  of  the  fraction 
by  the  complementary  surd,  V a + x + \/ a — x,  and  it 
becomes 


x{\/a  + x + Va  — x)  Va  + z + V a — x 

25  °r  2 

and  now  making  x — 0,  the  fraction  becomes  a/ a,  which  is 
its  true  value  when  x = 0. 


3. 


2x  — V 5x2  — a 2 
x — V2#2  — a 2 


when  x — a. 


Ans.  4. 


METHOD  OF  EVALUATION. 


105 


a — V a2  — z2 


when  x = 0. 


5.  -r , when  x = 1. 

x — 1 


Ans.  5. 


6.  Vz2  + ax  — x,  when  x = oo 


a 


There  are  many  indeterminate  forms  in  which  it  is  either 
impossible  to  detect  the  factor  common  to  both  terms,  or 
else  the  process  is  very  laborious,  and  hence  the  necessity  of 
some  general  method  for  evaluating  indeterminate  forms. 
Such  a method  is  furnished  us  by  the  Differential  Calculus, 
which  we  now  proceed  to  explain. 

METHOD  OF  THE  DIFFERENTIAL  CAL- 
CULUS. 

74.  To  evaluate  Functions  of  the  form 

Let  f(x ) and  <p  (x)  be  two  functions  of  x such  that 
f(x)  = 0 and  <p  (x)  = 0,  when  x = a. 


Let  x take  an  increment  h,  becoming  x -(-  h ; then  the 
fraction  becomes 


Now  develop  f(x  + h ) and  (p(x  + h)  by  Taylor’s  Theo- 
rem  ; substituting  h for  y in  (7)  of  Art.  66,  we  have 


Then  wTe  shall  have 


f(a)  _ 0. 


< p ( a ) 0 


f{z  + h) 
(f>  ( x + h) 


/(s  + h) 

<p{x  + h) 


f(x)  + f (x)  J + f"  (x)  ]l-  + etc. 


h 7/2 

0 (»)  + 0'  (x)  j + 0"  (x)  o + etc. 

-L  /v 


106 


METHOD  OF  EVAL UA TION. 


or  when  x — a, 


f (a  + h) 
<P  (a  + h) 


/(«)  +/'  («)  h +/"(«)  \ + etc. 

h2 

<p  (a)  + <p'  (a)  h + <p " ( a ) ^ + etc. 


(1) 


But  by  hypothesis  / (a)  = 0,  and  (p  (a)  — 0.  Hence, 
dropping  the  first  term  in  the  numerator  and  denominator, 
and  dividing  both  by  h,  we  have, 


/(„  + h)  _/»+/’» | + etc- 
0 («  + h)  ^ + 0"  (a)  | + etc. 


Now  when  = 0,  the  numerator  and  denominator  of 
the  second  member  become  /'  (a)  and  (p ' (a)  respectively  ; 
hence  we  have, 


./»  _ /'  (a) 
<p'  (ay 


<p  (a) 


/(*) 


as  the  7rt<e  value  of  the  fraction  -—7—  , when  x =z  a. 

<p  (x) 

(1.)  If  /”  (a)  = 0 and  <p'  (a)  be  not  0,  the  true  value  of 

/(«) 

0 («) 

(2.)  If  /'  (a)  be  not  aero  and  (p ' (a)  — 0,  the  true  value  of 

/ (a)  . 

(a) 


is  zero. 


is  OO. 


(3.)  If  /'  (a)  = 0,  and  <p'  (a)  = 0,  the  new  fraction 

is  still  of  the  indeterminate  form  Dropping  in 
</>  (a)  0 

this  case  the  first  two  terms  of  the  numerator  and  denomi- 

h2 

nator  of  (1),  dividing  both  by  -,  and  making  li  = 0, 
we  have. 


EXAMPLES. 


107 


/(«)  _ /"  0) 

0(a)  0(a)’ 

(#) 

as  the  true  value  of  the  fraction  — ; , when  % — a. 

0 (*) 

If  this  fraction  be  also  of  the  form  ^ , we  proceed  to  the 

next  derivative,  and  thus  we  proceed  till  a pair  of  deriva- 
tives is  found  which  do  not  both  reduce  to  zero,  when 
x = a.  The  last  result  is  the  true  value  of  the  fraction. 


EXAMPLES. 

1.  Evaluate  ^ ^ , when  x = 1. 
x — 1 

Here  / (x)  — log  x,  0 (t)  = x — 1 ; 

and  0'  (x)  = i ; 

1 

. f{x)  _ fjx)  _ X _ 1 _ J 

0 (a:)  0'  (a:)  — 1 ~~  a;Ji* 

That  is,  — 1 when  x — 1. 

x — 1 


2. 


Evaluate 


cos  a; 


when  x — 0. 


.n*) 

0'  (a;) 

rM 

0"  (a:) 


sin  a;  0 

~2aT  — O’ 


when  r = 0; 


cos  a: 


Therefore,  '^rr\ 

0 (®) 


1 


* The  subscript  denotes  the  value  of  the  independent  variable  for  which  the 
function  is  evaluated. 


108 


EVALUATION. 


X Sin  X — - 

S.  Evaluate , when  x = -• 

cos  x 2 


Here 


/'  (x)  x cos  x + sin  x 

(p'  (x)  — sin  x 


as 


5 — 1 


Hence  | — — 1. 

<P  {x) 

ax  — fc  . 

4.  — when  x = 0. 


(&& 

5.  7 7- > when  x — a. 

(x  — ay 


Ans.  log  -• 


Here 


/'  (z)  _ 1 _ 

<p'  (x)  ~ s(x  — a)'-lJa~ 


co  or  0, 


according  as  s > or  < 1. 


x — sin  x 

6.  - , when  x = 0. 


7. 

8. 

9. 


x? 

& — e~ 


2x 


x — sin  x 
e 1 — e~x 


, when  x = 0. 


, when  x = 0. 


sm  x 
e?  — 2 sin  x — e-1 


A ns.  f. 
Ans.  2. 
Ans . 2. 


x — sin  a; 


derivative. 


, when  x = 0.  Take  the  third 

Ajis.  4. 


(n2  ~2lf 

10.  when  x — a.  Cancel  the  factor  (a  — x)l 


(a  — x)? 


Ans.  (2a)^c 


75.  To  evaluate  Functions  of  the  form 

f(x)  _ QO_ 

<p  (x)  OO  ’ 


CO 

00 


Let 


when  x = a. 


EVALUATION. 


109 


Since  the  terms  of  this  fraction  are  infinites  when  x = a, 
their  reciprocals  are  infinitesimals  (Art.  8)  ; that  is, 

= 0,  and  — i-r  = 0,  when  x — a ; hence, 

*(x)  0 ( x ) 


1 

fix)  _ <P  (x)  _ 0 
<p  (x)  1 0 ’ 


fix) 


when  x = a, 
1 


and  therefore  the  true  values  of 


0 (x) 

1 


may  be  obtained  by 


f(x) 

Art.  74 ; that  is,  by  taking  the  derivatives  of  the  terms, 
thus, 


1 0'  (x) 

fjx)  _ 0jA)  _ [0  (s)]8  _ 0'(g)  [/(s)]8 

0 ix)  _j_  f ix)  fwivi*)]*’ 

f(x)  Uix)Y 


when  x = a, 


nr  /(£)  _ 0'  («)  [/(«)? 

0 (a)  f («)  [0  («)]*’ 

Dividing  by  |^-j,  we  get, 

= 0'  (fl)  _ fja ) , 

f («)  ’ 0 («)  ’ 


whence 


fja)  __ /' («) 

0(a)  0'(«)’ 


Hence  the  true  value  of  the  indeterminate  form  — • is 

CO 


found  in  the  same  manner  as  that  of  the  form 


0 

o' 


In  the  above  demonstration,  in  dividing  the  equation  bv  when 

f («) 

x = a,  we  assumed  that  l is  neither  0 nor  oo , so  that  the  proof 
<p{a) 

would  fail  in  either  of  these  cases. 


110 


EXAMPLES. 


It  may,  however,  he  completed  as  follows  : Suppose  the  true  value 

of  — L\  to  be  0 ; then  the  value  of  ~/(a)  + jg  ^ where  h may 

0 («)  0 (a) 

he  any  constant.  But  as  this  latter  fraction  has  a value  which  is 

f '(a)  + h <p' (a) 


neither  0 nor  go,  its  value  by  the  above  method  is 


0'(«) 


+ h ; and  since  the  value  of  this  fraction  is  h,  the  first  term 


_ f («) 

0'  (*) 

where  ^ = 0,  OS  is  also  0. 

0 (a)  b (a)  0 (a) 

f(x) 

Similarly,  if  the  true  value  of  J—7-[  be  oo  when  x — a.  then 

0(z) 

= 0 : and  therefore  we  have  - = 0,  by  what  has  just  been 

i f (a) 

shown  ; /.  - — . - = oo. 


0(a) 


/’(«) 


Therefore,  in  every  case  the  value  of  - determines  the  value  of 
J 0 («) 

— - for  either  of  the  indeterminate  forms  --  or  — . (See  Williamson’s 
0 (a)  0 oo  v 

Dif.  Cal.,  p.  100.) 


EXAMPLES. 

1.  Evaluate  -3°  X- , when  x = oo . 
xn 


Here  ^ = A 

< I > (a;)  </>'  (a:)  ?ia;n_I 


1 0 o'  ^ 

2.  Evaluate  — , when  a;  = 0. 
cot  a: 


nxn 


= 0. 


Here 


f (X) 

x 

sin2  x 

<p'  (*) 

— cosec2  a:  - 

x _ 

f"  (*) 

2 sin  x cos  x~ 

= 0; 

0 

0"  (*)  “ 

1 

0, 

0’ 


X = 0,  when  x = 0. 
cot  x 


O (O 


EVALUATION. 


Ill 


1 — log  X 


X 

ex 


, when  x — 0. 


n 

4:X 

. TTX 
Got  -q- 


, when  x = 0. 


_ log  tan  CZx)  , 

5.  — - — - — when  x = 


log  tan  x 


Ans.  0. 


Ans. 


TT2 

8 ' 


Ans.  1. 


76.  To  evaluate  Functions  of  the  form  0 x co . 

Let  f(x)  x (p  (x)  = 0 x co , when  x = a. 

The  function  in  this  case  is  easily  reducible  to  the  form 
for  if  f (a)  = 0,  and  <p  (a)  — oo,  the  expression  can  be 

written  — y- , which  = therefore  f(x)  x <p(x)  = 

<p  (a)  <p{x) 

may  be  evaluated  by  the  method  of  Art.  74. 


EXAMPLES. 


TTX 


1.  Evaluate  (1  — x)  tan  — , when  x = 1. 


We  may  write  this 


Here 


/'(*) 

<p'(x) 


1 — X 
, TTX 

cot  -JT- 


— 1 

TT  TTX 

— - cosec2  — 
Z 'I 


2 sin3  (f ) 


_1  tt 


2.  Evaluate  xn  log  x,  when  x — 0. 

1 


Xn  log  X — 


log  X 
xrn 


— war 


sf1 

n 


= 0. 


112 


EXAMPLES. 


3. 

4. 


e~*  log  x,  when  x = go  . 


sec  x (x  sin  x — when  x = 


^4w.s.  0. 
.4ks.  — 1. 


77.  To  evaluate  Functions  of  the  form  co  — co . 

Let  f(x)  and  0 (a;)  be  two  functions  of  x which  become 
infinite  when  x = a.  Then  f (x)  — 0 (x)  = go  — co , when 
x — a. 

The  function  in  this  case  can  be  easily  reduced  to  the 
form  - , and  may  be  evaluated  as  heretofore. 


EXAMPLES. 


1.  Evaluate 


2 


, when  x = L 


x 4 — 1 x — 1 

This  takes  the  form  co  — oo , when  x = 1. 

2 1 2 — a;  — 1 0 

-5 — x- — x—  = xr , when  x = L 

ar2  — 1 a:  — 1 a;2— 1 0 


/'(*)  -1  , 

0'  (a;)  2a: 


when  a:  = L 


2.  Evaluate  — — r — ^0r~  — , when  a;  c=  0 

a:  (1  + a:)  x 2 

iphich  takes  the  form  00  — 00 , when  x = 0. 

1 log  (1  + z)  _ g — (1  + x)  log  (1  + x) 

z(l4-a;)  xl  ~ a^(l+a:) 

_ x — (1  + x)  log  (1  4-  X ) 

a^ 

(remembering  that  14-2  = 1,  when  x vanishes). 


EVALUATION. 


113 


f{x)  1 — log  (1  + x)  — 1 0 

<&'  (x)  2x  O' 


&- 

<p"{x)  - 


1 + X 

2 _ o 


= ~b 


when  x = 0 ; 


3.  Evaluate  sec  x — tan  x,  when  x = -• 

Z 

1 — sin  x 0 . 7 r 

sec  x — tan  x — = - , when  x — 

cos  a:  0 2 

f'(x)  _ - cos  aTl  _ Q 

< p ' (x)  — sin  x\  t 


Hence,  sec  - and  tan  - are  either  absolutely  equal,  or 
Z Z 

differ  by  a quantity  which  must  be  neglected  in  their  alge- 
braic sum.* 


X 

1 

when  x = 1. 

Ans. 

X — 1 

log  x' 

1 

log  X 

X 

log  a;’ 

when  x = l. 

A ns.  — 

78.  To  evaluate  Functions  of  the  forms  0°,  co°, 
and  1±0°. 

Let  f(x)  and  <p  (x)  be  two  functions  of  x which,  when 
x = a,  assume  such  values  that  [/(x)]^te)  is  one  of  the 
above  forms. 

Let  y = [/«]*w: 

•••  log  y - 0 (a?)  log /(a;), 

( 1 .)  When  f(x)  — oo  or  0,  and  (f>  (x)  = 0. 

log  y = 0 (x)  log / (x)  - o (±  oo  ), 
which  is  the  form  of  Art.  76. 


* Price’s  Infinitesimal  Calculus,  Vol.  I,  p.  210. 


114 


EXAMPLES. 


Hence,  [/(z)]*^  becomes  indeterminate  wlien  it  is  ol 
the  form  0°  or  oo°. 

(2.)  When  f (x)  = 1,  and  0 (x)  = ± oo. 

lo  gy  = <P  (x)  lo  gf(x)  = ±cc  x 0. 

Hence,  [/(z)]0u)  is  indeterminate  when  of  the  forms 

CO 

Hence  the  indeterminate  forms  of  this  class  are 
0°,*  oo  o,  1±0°, 

and  may  all  be  evaluated  as  in  Art.  76,  by  first  evaluating 
their  logarithms,  which  take  the  form  0x«. 


EXAMPLES 

1.  Evaluate  xx,  when  x — 0. 


We  have 


hence, 


log  xx  = x log  x = 


kg  x. 

XT'  * 


1 


/' (s)  X 

0'  ( X ) — x~ 2 

log  a^  = 0,  when 
a*  = 1,  when 


— x]9  = 0; 

x = 0; 
x = 0. 


2.  Evaluate  xf,  when  x = oo. 

i 4 1,  log  a; 

logaf  = - log  x = — ; 

1 

/'  fo)  _ «_  _ l"l  _ 0. 

*'(*)“  i “d.”  ’ 


* In  genera],  the  value  of  the  indeterminate  form  0°  is  1.  (See  Note  on  Inde- 
terminate Exponential  Forms,  hy  F.  Franklin,  in  Yol.  I,  No.  4,  of  American  Journal 
of  Mathematics.) 


EXAMPLES. 


U5 


log  af  — 0,  Avhen  x — oo  ; 
hence,  af  = 1,  when  x = go  . 


3.  Evaluate 


(i+tr. 


when  a;  = =o  . 


Let  x — -,  and  denote  the  function  by 
z 7 

Then  m = (1  + az)z\ 

(since  when  x = qo  , z = 0)  ; and 

log  u = ^ + aZ ^ , when  ? = 0. 

° z 

Taking  derivatives,  we  have 

i a 

log  tia  - = a : 

1 + az  Jo 

log  (l  + ~)  = a>  when  x — go  ; 

(g\X 

1 + - j — ea,  when  a:  = <x> . 

If  a — 1,  we  have 

(i + !}■  = .« 

CO 

that  is,  as  a:  increases  indefinitely,  the  limiting  vcihie  (Art 
41)  of  the  function  ^1  + ^ is  the  Naperian  base. 


*•  !)“'• 

5.  afin% 


/ . 7TJ> 


when  a;  = 0. 
when  z = 0. 
when  x — a. 


Am.  1. 

^4  ns.  1. 

£ 


116 


COMPOUND  INDETERMINATE  FORMS. 


79.  Compound  Indeterminate  Forms. — If  an  inde- 
terminate form  be  the  product  of  two  or  more  expressions, 
each  of  which  becomes  indeterminate  for  the  same  value  of 
x,  its  true  value  can  be  found  by  evaluating  each  factor 
separately;  also,  when  the  value  of  any  indeterminate  form 
is  known,  that  of  any  power  of  it  can  be  determined. 


EXAMPLES. 


1.  Evaluate  — , when  x = <x>. 
ex 

This  fraction  may  be  written  {^7^  * 

X 

We  first  evaluate  — , when  x = go  . 
e » 


Here 


Hence, 


f'W)  _ J_- 
(*)  l ^ 

n 


= — = 0. 

00 


S'  = o-  = o. 

e* 


2.  Evaluate  xm  log"  x,  when  x = 0,  and  m and  n are 
positive. 


Here 


(x"  log  x)n  — 


= (fey. 


IqOT  ‘X 

We  first  evaluate  8 , when  x — 0. 


X n 


W 3 have 


/'(*)  _ 

0'  (*) 


W — - — 1 

a;  • 

n 


n - 
m 


s'"  log"  x — un  — 0. 


= 0. 


EXAMPLES. 


117 


i-v-TYl  /-pTr\-\-fl 

3.  — ; 5— , when  x = 1. 

1 — x?p 

This  function  can  be  written  in  the  form 

xm  1 — xn 

1 + XP  ' 1 — xp' 

We  have  to  evaluate  only  the  latter  function  for  x = 1, 
since  the  former  is  determinate. 


Here 


/'(a)  _ — nx 
(p1  (x)  — px 


n 


XT  1 

1 + XP  ~ 2’ 

1 — x -P  ~~  2 p ’ 


P— 1 


n 

ryAl — f) 

~ P 


when  x = 
when  x — 
when  x = 


1. 

1. 

1. 


(x2  ~ «2)  sin  x- 

4. — , when  x = a. 

9 TO 

X 2 cos  — 

2 a 


— a2)  sin  — 


. TO 

x2  — a 2 Sin  2a 


x2  cos 


TO 

2a 


cos 


TO 

2a 


x 


2 * 


We  have  only  to  evaluate  the  first  factor, 


x2  — aF~ 


cos 


nx 

2a 


2x 


n . nx 
— — sin  ^ 
2a  2a 


4 a2 


. nx 
sin  — 
2 a 


and 


_i  a 


118 


EXAMPLES. 


( x 2 — a2)  sin : 


TXX 

2a 


X 2 cos 


2a 


EXAMPLES. 


1. 


ex  — e x 
log  (1  + x)  ’ 


x3  — 5a*2  + 7x  — 3 

2.  — s , when  x = 3. 


x 3 — x 2 — ox  — 3 
3.  (siL»f )-,  when  x = 0 

ex  — e~x  — 2x 


when  x = 0.  Ans.  2. 

b 


nr. 


4. 


(ex  — l)a 


— (dif.  three  times),  when  x = 0. 

b 

1. 


_ 1 — sin  x -f  cos  x , t t 

5.  - , when  x = -• 

sm  x + cos  x — 1 2 


6. 


tan  x — sin  a: 
sin3  x 

a8'" x — a 


8. 


log  sin  x ’ 

x2  + 2 cos  x — 2 
tf— 


when  x — 0. 


when  x = x* 

-v 


a logo. 


(dif.  four  times),  -when  2 = 0. 


9.  a:1  1 (pass  to  logarithms),  W'hen  x — 1. 

x log  (1  + x) 


10 


1 — cos  X 


■when  x = 0. 


II.  x-e1,  when  x — 0. 


ao. 


EXAMPLES. 


119 


12.  (pass  to  logarithms),  when  x = 


00  . 
A ns.  1. 


13.  2x  sin  — , when  x = co . a. 

2x 

14.  ex  sin  x,  when  x = 0.  oo . 

15.  (cos  ai)cosec*cx  (pass  to  logarithms,  and  dif.  twice), 
tvhen  x — 0. 


e 


16.  (sin  xy™*  (x~r~~x-)  (see  Art.  79),  when 
sm  axj 


17.  (sin  x)tanar,  when  !&  — -• 


7 Tm 

L 


iNil  3 


CHAPTER  VII. 


FUNCTIONS  OF  TWO  OR  MORE  VARIABLES,  AND 
CHANGE  OF  THE  INDEPENDENT  VARIABLE. 

80.  Partial  Differentiation. — In  the  preceding  chap- 
ters, we  have  considered  only  functions  of  one  independent 
variable:  such  functions  are  furnished  us  in  Analytic 
Geometry  of  Two  Dimensions.  In  the  present  chapter,  we 
are  to  consider  functions  of  two  or  more  variables.  Analytic 
Geometry  of  Three  Dimensions  introduces  us  to  functions  of 
the  latter  kind.  For  example,  the  equation 

z = ax  A by  + c (1) 

represents  a plane ; x and  y are  two  independent  variables, 
of  which  z is  a function.  In  this  equation,  z may  be 
changed  by  changing  either  x or  y,  or  by  changing  them 
both,  as  they  are  entirely  independent  of  each  other,  and 
either  of  them  may  be  considered  to  change  without  affect- 
ing the  other ; in  this  case  z,  the  value  of  which  depends 
upon  the  values  of  x and  y,  is  called  a function  of  the  inde- 
pendent variables  x and  y. 

A partial  differential  of  a function  of  several  variables  is 
a differential  obtained  on  the  hypothesis  that  only  one  of 
the  variables  changes. 

A total  differential  of  a function  of  several  variables  is  a 
differential  obtained  on  the  hypothesis  that  all  the  variables 
change. 

A partial  derivative  of  a function  of  several  variables  is 
the  ratio  of  a partial  differential  of  the  function  to  the  dif- 
ferential of  the  variable  supposed  to  change. 


PARTIAL  DIFFERENTIATION. 


121 


A total  derivative  of  a function  of  several  variables  is  the 
ratio  of  the  total  differential  of  the  function  to  the  differen- 
tial of  some  one  of  its  variables.  (See  Olney’s  Calculus, 
p.  45.) 

As  all  the  variables  except  one  are,  for  the  time  being, 
treated  as  constants,  it  follows  that  the  partial  differentials 
and  derivatives  of  any  expression  can  be  obtained  by  the 
same  rules  as  the  differentials  and  derivatives  in  the  case  of 
a single  variable. 

If  wre  differentiate  (1),  first  with  respect  to  x.  regarding 
y as  constant,  and  then  with  respect  to  y,  regarding  x as 
constant,  we  get 

dz  = adx,  (2) 

and  dz  — bdy.  (3) 


Dividing  (2)  and  (3)  by  dx  and  dy  respectively,  we  get, 


II 

a 

(4) 

and 

£1* 

II 

(5) 

The  expressions  in  (2)  and  (3)  are  called  the  partial 
differentials  of  2 with  respect  to  x and  y,  respectively,  while 


— and  -7~-  are  called  the  partial  derivatives  of  z with  re- 
dx  dy 

spect  to  the  same  variables. 

Since  a and  b in  (4)  and  (5)  are  the  partial  derivatives  of 
z with  respect  to  x and  y,  respectively,  we  see  from  (2) 
that  the  partial  differential  of  z with  respect  to  x,  is  equal 
to  the  partial  derivative  of  z with  respect  to  x multiplied 
by  dx,  and  similarly  for  the  partial  differential  of  y. 

Hence,  generally,  if 

/(*>  V’  z ) 

denotes  a function  of  three  variables,  x,  y,  z,  its  derivative 
or  differential  when  x alone  is  supposed  to  change,  is  called 

6 


122 


PAR T1AL  DIFFERENTIA  TION. 


the  partial  derivative  or  differential  of  the  function  with 
respect  to  x,  and  similarly  for  the  other  variables,  y and  z. 
If  the  function  be  represented  by  u,  its  partial  derivatives 
are  denoted  by 

du  du  du 

dx’  dy ’ dz’ 


and  its  partial  differentials  by 

du 


du  7 

aidx’ 


dy 


dy, 


du 

dz 


dz. 


81.  Differentiation  of  a Function  of  Two  Varia- 
bles.— Let  u — f (x,  y),  and  represent  the  partial  differ- 

du 

ential  of  u with  respect  to  x,  by  — dx,  and  with  respect 
du 

to  y , by  — dy,  while  du  represents  the  total  differential. 

Let  x and  y receive  the  infinitesimal  increments  dx  and 
dy,  and  let  the  corresponding  increment  of  u be  du.  Then 
we  have, 

du  = f{x  + dx,  y + dy)  — f{x,  y). 

Subtract  and  add  f (x,  y + dy),  and  we  have 
du  — f (x + dx,  y + dy)  —fix,  y+dy)  +f(x,  y + dy) 

—f{z>  y )• 

du 

Now  f(x  + dx,  y + dy)—f(x,  y + dy)  = ^ dx,  because 

it  is  the  difference  between  two  consecutive  states  of  the 
function  due  to  a change  in  x alone;  that  is,  whatever  dif- 
ference there  is  between  f(x  + dx,  y + dy)  and  f(x,  y + dy) 
is  due  solely  to  the  change  in  x,  as  y + dy  is  the  value  of  3 
in  both  of  them.  For  the  same  reason, 

f(x,  y + dy)  — / (x,  y)  = ~^dy; 

, du  , du  7 

du  = TxdX  + ^iy, 


and  therefore  we  have 


EXAMPLES. 


123 


dw  da 

in  which  dx  and  dy  are  the  partial  differentials  of  u 

with  respect  to  x and  y,  respectively,  while  du  is  the  total 
differential  of  u when  both  x and  y are  supposed  to  vary. 
In  the  same  way,  we  may  find  the  differential  of  any  num- 
ber of  variables. 

Hence,  the  total  differential  of  a function  of  two  or  more 
variables  is  equal  to  the  sum  of  its  partial  differentials. 

The  student  will  carefully  observe  the  different  meanings 
given  to  the  infinitely  small  quantity  du  in  this  equation, 
otherwise  the  equation  will  seem  to  be  inconsistent  with  the 

da 

principles  of  algebra.  Thus,  in  dx,  du  denotes  the  in- 
finitely small  change  in  u arising  from  the  increment  dx  in 
x,  y being  regarded  as  constant.  Also,  in  ^ dy,  du  denotes 

the  infinitely  small  change  in  u arising  from  the  increment 
dy  in  y,  x being  regarded  as  constant,  while  du  in  the  first 
member  denotes  the  total  change  in  u caused  by  both  x and 
y changing.  If  the  partial  differentials  of  x and  y be  rep- 
resented by  dxu  and  dyu,  respectively,  the  preceding  equa- 
tion may  be  written 

du  = d,u  + d,u. 


EXAMPLES. 

1.  Let  u — ay 2 -f  bxy  + cx 2 + ey  + gx  + h,  to  find  the 
total  differential  of  u. 

Differentiating  with  respect  to  x,  we  have 
dxu  = bydx  + 2 cxdx  + gdx. 

Differentiating  with  respect  to  y,  we  have 
dyu  — 2aydy  -f-  bxdy  -f  edy. 

Hence,  du  — (by  A 2 cx  -f  g)  dx  + (2 ay  + bx  + e)  dy. 


124 


EXAMPLES. 


2.  U - 
Here 

Hence, 

3.  u - 
Here 
Hence, 

4.  u : 

Here 

Hence, 

5.  u - 
Here 

Hence, 

6.  u - 


Xy. 


dxu  = yxy  1 dx,  d,,u  = x)>  log  x dx 
du  — yxy~l  dx  + xy  log  x dy. 


x 2 y2 

a2  + b2 


dxu  — ^ dx.  d u = ~dy 
a i l2 

du  = — dx  + ^ dy. 

a 2 b2  * 


tan-1  V~- 

x 


dxu 


ydx 

x2  ydx 


1 + 


y* 


duu  = 


dy 

x 


x2  -f  y 2 


xdy 


du  = 


w2  X2+  f 
+ X2 

xdu  — ydx 
x?  + y2 


sin-1  - + sin  , 
a b 


-i  y. 

b 

dx  , dy 


dxu  = — — — , dyu  = , 

¥ a2  — a?  Vb2  — y* 


du  = -JL=  + 


dy 


V a2  — x2  Vb2  — y2 

t/sln  *.  du  — y,m  1 log  y cos  x dx  + 


sin  x dy 

ycoren  * 


TOTAL  DERIVATIVE. 


125 


7.  u = vers-1  -• 

y 

8.  u — log 


du  — ~ X(^y 

yVftxy  — x 2 

V 

du  — ^dx  + log x dy. 


82.  To  Find  the  Total  Derivative  of  u with  re- 
spect to  x,  when  u = / (y,  z),  and  y — <j>  {x), 
z — 0i  (as). 

Since  u =f{y,  z),  we  have  (Art.  81), 

T du  7 du  , 

and  since  y = <p(x),  we  have  dij  = tp-dx', 

cLz 

since  z — <Pi  (x),  we  have  dz  — — dx. 


Substituting  these  values  for  dy  and  dz  in  (1),  wre  get 


, du  du  , du  dz  7 
du  = -=-  -r-  dx  + -r-  -r-  dx. 
dy  dx  dz  dx 


(2) 


Dividing  by  dx,  and  denoting  the  total  derivative  by  ( ), 
we  have 


(du\  du  dy  du  dz 

dx)  dy  dx  dz  dx 


(3) 


Cor.  1. — If  z = x,  the  proposition  becomes  u = f{x,  y) 
dz 

and  y — (p  (x);  and  since  1,  (3)  becomes 


tdu\  _ du  du 
\dx)  ~ dx'  dy 


dx 

du  du  dy 


dy  dx 

Cor.  2. — If  u —f{x,  y,  z),  and  y = <p  {x),  and  z = <f) , ( x ), 
we  have 

, du  7 du  7 du  7 

iu  = Txdx+Tyd^<hdl- 


(1) 


126 


EXPLANATION  OF  TERMS. 


dy  = d£dx, 


and 


dz  = ^ dx. 
dx 


Substituting  the  values  of  dy  and  dz  in  (1),  and  dividing 
by  dx,  we  get 


_ du  ^ du  dy  du  dz 
dx  dy  dx  dz  dx 


Cor.  3.  — If  u = f (y,  z,  v),  and  y — <p  (x),  and  z = 
<pi  (x),  and  v — <p2  (x),  we  have, 


, du  , du  . dv,  , 
du  — dy  + -r-  dz  + — dv. 
dy  J dz  dv 


(1) 


dy  — ^~.dx',  dz  = ^ dx 


dv  = ~ dx. 
dx 


Substituting  the  values  of  dy,  dz,  dv,  in  (1),  and  dividing 
by  dx,  we  get 

/du\  du  dy  du  dz  du  dv 

\dx)  dy  dx  + dz  dx  dv  dx 


Cor.  4. — If  u — f{y)  and  y — <t>  (x),  to  find 


du 

dx 


du 

Since  u — f{y)>  have  du  = — dy. 

Since  y = <p  (x),  we  have  dy  — ^ dx. 

7 du  dy  7 7 du  dudy 

therefore,  du  — dx,  and  -r  = -r  -f1- 

dy  dx  dx  dy  dx 


Sch. — The  student  must  observe  carefully  the  meanings 
of  the  terms  in  this  Art.  Thus,  in  the  Proposition,  u is 
indirectly  a function  of  x through  y and  z.  In  Cor.  1,  u is 
directly  a function  of  x and  indirectly  a function  of  x 
through  y.  In  Cor.  2,  u is  directly  a function  of  x and 


EXAMPLES. 


127 


indirectly  a function  of  x through  y and  z.  In  Cor.  3,  u is 
indirectly  a function  of  x through  y,  z,  and  v.  In  Cor.  4, 
u is  indirectly  a function  of  x through  y. 

The  equations  in  this  Article  may  seem  to  be  inconsistent 
with  the  principles  of  Algebra,  and  even  absurd  ; but  a little 
reflection  will  remove  the  difficulty.  The  du' s must  be 
carefully  distinguished  from  each  other.  In  Cor.  1,  for 
du 

example,  the  du  in  — is  that  part  of  the  change  in  u 

which  results  directly  from  a change  in  x,  while  y remains 
du 

constant ; and  the  du  in  — is  that  part  of  the  change  in  u 

which  results  indirectly  from  a change  in  x through  y ; and 

the  du  in  is  the  entire  change  in  u which  results 

directly  from  a change  in  x,  and  indirectly  from  a change 
in  x through  y. 

EXAMPLES. 


1. 


u = tan  -1  - and  y = (r2  — x2)?,  to  find 


Here 


du  y du 

dx  ~ r2’  dy 


x 

^2’ 


and  ~ = — x> 
dx  y 


Substituting  in  (|)  = g + g g (Art.  83.  Cor.  1),  we 
have  3 


y2  + x2  _ 1 

r^y  vV2  — x2’ 


and  this  value  is  of  course  the  same  that  we  would  obtain 
if  we  substituted  in  u — tan-1  - for  y its  value  in  terms  ol 

y 

x,  and  then  differentiated  with  respect  to  x. 


128 


EXAMPLES. 


2.  u = tan-1  (xy)  and  y = ex,  to  find  • 


du  _ 

dx  ~ 1 + 2%2’  dy  ~ 
(Art.  82,  Cor.  1), 

(du\  y + exx 
\dx)  ~ T 


1 + x^y2’  dx 
_ e®(l4-z)  _ 


= e*. 


4-  x2y2  1 + a^e2*  ’ 

and  this  value  is  of  course  the  same  that  we  would  obtain  ii 
we  differentiated  tan-1  ( xex ) with  respect  to  x. 

3.  u — z2  + y3  + zy  and  z = sin  x,  y = e®,  to  find 


Here 


du 

dy 

dz 


— 5^/2 


dx 

(Art.  82), 


= cos  xt 


du 

dz 


= %z  + y, 


k-<* 

dx  ' 


(£)  = ^ + *)  e*  + (2*  + y)  cos  a; 

= (Se2®  + sin  x)  e®  + (2  sin  z + e®)  cos  x 
= 3c3®  + e®  (sin  a:  + cos  x)  + sin  2x. 

(See  Todhunteffs  Dif.  Cal.,  p.  150.) 

Let  the  student  confirm  this  result  by  substituting  in  u 
for  y and  z,  their  values  in  terms  of  x,  thus  obtaining 

u — e3®  4-  e®  sin  x + sin2  x, 

and  then  differentiate  with  respect  to  x. 

4.  u — sin-1  (y  — z),  y — 3x,  z — hr3. 
du  _ 1 

dy  Vl  — (y  — z)* 


EXAMPLES. 


129 


du 

dz 


® = 3, 


dx 

tdu 


= 12a3. 


Vl  — (y  — *)2 

dz 
dx 

, (Art. 82),  (^)  = — -~-12-*L 

'^*7  a/i  — (^/  — z)2 

3 — 12a^ 


eax  (y  — z)  , 

5.  « = ^ and 

a1  -f-  1 


Vl  — 9a?  + 24a,4  — ] 6a?  Vi  — a;2 

y — a sin  x,  z = cos  a:. 


du 

eox 

du 

eox 

dy 

a2  + 1’ 

dz 

a2  + V 

II 

5sj|  H 
p53  I'SS 

a cos  x, 

dz 

dx 

— sin  x. 

c/m  «ea*  (y  — z) 

dx  — a2  + 1 

(Art.  82,  Cor.  2), 

(du\  e0*  , . „ . . 

I ~r  I = -5 ( a cos  x + sin  x 4-  a2  sm  x — a cos  x) 

\dxl  a2  + 1 v ’ 

— e0*  sin  x.  (See  Courtenay’s  Cal.,  p.  73.) 

6.  u = yz  and  y — e*,  z — a?  — 4a?  + 12a:2  — 24a:  + 24, 

(S)  = ~ 

7.  If  u =f  (2)  and  z = <p(x,  y),  show  that 

, du  dz  7 du  dz  , 
du  = -r-  -rr-  dx  + — -y-  dy. 

rl~  dz  dy  17 


dz  dx 

a x4y2  xiy  xi  ^ , 

8.  u — + 32  and  y = log  x. 


(du\  _ 
dx)~ 


= x?  (log  a:)2. 


130 


PARTIAL  DIFFERENTIATION. 


83.  Successive  Partial  Differentiation  of  Func- 
tions of  Two  Independent  Variables. 


Let 


u =f(x>  y) 


be  a function  of  the  independent  variables  x and  y\  then 

da  du  . -i  p <•  n . 

and  — are,  m general,  functions  or  x and  y,  and  hence 

may  be  differentiated  with  respect  to  either  x or  y,  thus 
obtaining  a class  of  second  partial  differentials.  Since  the 
partial  differentials  of  u with  respect  to  x and  y have  been 
el'll/  du 

represented  by  dx  and  — dy  (Art.  81),  we  may  repre- 


sent the  successive  partial  differentials  as  follows : 

The  partial  differential  of  dxj,  with  respect  to  x, 
d tdn  7 \ , 

= Tx\x/x)ix' 


which  may  be  abbreviated  into 


dx* d ’ 


The  partial  differential  of  dxj,  with  respect  to  y, 


d /du  7 
= Ty\Txix 


) ,]>J, 


which  may  be  abbreviated  into 


cPu 
dy  dx 


dy  dx. 


rJ2  n jJZfi 

Again,  both  | _;7  dx*  and  (j}-jx  dy  dx  will  generally  be 

functions  of  both  x and  y,  and  may  be  differentiated  with 
respect  to  x or  y,  giving  us  third  partial  differentials,  and 
60  on.  Hence  we  use  such  symbols  as 


PARTIAL  DIFFERENTIATION. 


131 


(P'U 
dy  dx2 


dy  dx2, 


d?U 

- — t—z-  dx  dy  dx,  and 
dx  dy  dx 


d?u 
dy 2 dx 


dy 2 dx, 


the  meaning  of  which  is  evident  from  the  preceding  re- 

(fill 

marks.  For  example,  - — =— — dx  dy  dx  denotes  that  the 

(X/JO  tl  if  Cl  JO 

function  u is  first  differentiated  with  respect  to  x,  supposing 
y constant;  the  resulting  function  is  then  differentiated 
with  respect  to  y,  supposing  x constant;  this  last  result  is 
then  differentiated  with  respect  to  x,  supposing  y constant; 
and  similarly  in  all  other  cases. 

When  u — f (x,  y),  the  partial  derivatives  are  denoted  by 
d2u  d2u  d2u  d3u  d3u  d3u 
dx2  ’ dy2  ’ dx  dy  ’ dx? ' dx2  dy  ’ dx  dy2  ’ 6 C' 


84.  If  u be  a Function  of  x and  y,  to  prove  that 


cPu  , , d2u  , , 

-= — dx  dy  = , — ? - dy  dx. 
dx  dy  dy  dx 


Take 


du 

dx 


u — x2y3, 


dx  — 2xy3dx , 


jy  = ox2y2dy, 


d2u 


dy  dx  — Qxy2  dy  dx, 

(fin. 

-= — 7—  dx  dy  — Qxy2  dx  dy. 
dx  dy  s s 3 


In  this  particular  case, 
d2u 


7 , dx  dy  = fiU1  ■ dy  dx ; 
dx  dy  3 dy  dx  3 


that  is,  the  values  of  the  partial  differentials  are  independ- 
ent of  the  order  in  which  the  variables  are  supposed  to 
change. 


132 


PARTIAL  DIFFERENTIATION. 


To  show  this  generally : 

Let  u — fix,  y); 

then  jx dx  = f(x+dx>  y}  —f(x>  V)' 

This  expression  being  regarded  as  a function  of  y,  let  y 
become  y + dy,  x remaining  constant ; then 

d (du  , 

•=-(  -^-dx 
cly  \dx 


) dV  — f(x  + dx,  y + dy)  —f{x,  y + dy) 

— (f(x+dx,y)—f{x,y)] 
— f(x  + dx,  y + dy ) — f(x,  y + dy) 

—f(x  + dx,y)+f(x,y). 

In  like  manner, 


~dy  =f{x,  y + dy)  -f(x,  y). 


Jx  (£  dy)  dX  =f(x  + dx>  y + dd)  f(x+dx,  y) 

J -[f(+y+dy)-fi+y)] 

= f(x  + dx,  y + dy)  —f(x  + dx,  y) 

— / (x>  y + dy)  +f{x,y). 

These  two  results  being  identical,  we  have 

Tv\Txax)dy  = TxWy)ix ’ 


that  is, 


d2u  7 7 d2u  j j 

- — — dy  dx  = y dx  dy. 

dy  dx  3 dxdy 


Dividing  by  dy  dx,  we  get 

cPu  <Pu 


dy  dx  dx  dy 

In  the  same  manner,  it  may  be  shown  that 

dhi  d?u 

dx?  dy  ~ dy  do. 9* 
and  so  on  to  any  extent. 


or 


EXAMPLES. 


133 


EXAMPLES. 


1.  Given  u = sin  (x  + y),  to  find  the  successive  partial 
derivatives  with  respect  to  x. 

xz  - cos  (*  + y)>  = - sm  (*  + y)> 


dsu 

dx? 


— cos  (x  + y), 


d^u  . , . 

-Xa  = sin  (*  + ?/)’  etc* 


2.  m = log  (a;  + y),  to  find  the  successive  partial  de- 
rivatives with  respect  to  x,  and  also  with  respect  to  y in 
the  common  system. 


du 


m 


d?u 
dx 2 


m 


. (x -\ - y)2’ 
m 

~ + y)2’ 


dx  x + y 

du  m d2u 

dy  ~ x + y ’ dy2 

(See  Art.  65,  Lemma.) 

3.  If  u — x log  y,  verify  that 

4.  If  u = tan-1  verify  that 

5.  If  u — sin  (axn  + hyn), 

verify  that 


d3u  2 m 


dx 3 

(x  + y)3 

d3u 

2 m 

dy3  ~ 

~ (*  + yf 

d2u 

cPu 

dydx 

- dxdy 

d*u 

d3u 

dy2dx 

~ dxdy2 

d*u 

diu 

dx2dy2  dy2dx 2 


85.  Successive  Differentials  of  a Function  of  Two 
Independent  Variables. 

Let  u —f  (x,  y). 

We  have  already  found  the  first  differential  (Art.  81), 


134 


SUCCESSIVE  DIFFERENTIALS. 


Differentiating  this  equation,  and  observing  that  — — 

dx’  di/' 

are,  in  general,  functions  of  both  x and  y (Art.  83),  and 
remembering  that  x and  y are  independent,  and  hence 
that  dx  and  dy  are  constant,  we  have, 


d?u  = 


(sdx)  d[fxdx) , d(%dy) 

—d~ dx  + —iy-  d'J  + — lr-  * 


cPu  7 , , (Pu  , 7 d?u  7 7 d?u  7 , 
^ ^ ^ 


cfe2 

d?u 
dx 2 


-S<w  + *a^*4r  + ||^. 


(2) 


(fin,  cfiit 

(since  efycfa;  = dxdy,  Art.  84). 


Differentiating  (2),  remembering  that  each  term  is  a 
function  of  a;  and  y,  and  hence  that  the  total  differential 
of  each  term  is  equal  to  the  sum  of  its  partial  differentials, 
we  get, 


= + + »=?**•+$*•■<*> 

and  so  on.  It  will  be  observed  that  the  coefficients  and 
exponents  in  the  different  terms  of  these  differentials  are 
the  same  as  those  in  the  corresponding  powers  of  a bino- 
mial ; and  hence  any  required  differential  may  be  written 
out. 


* The  total  differential  of  each  of  the  terms 

sum  of  its  partial  differentials. 


(“”  tfz)  and  (^  d y)  is  equal  to  the 


IHP  LICIT  FUNCTIONS 


135 


EXAMPLES. 


1.  u = (x2  + y2)k 


du 

dx 


(x*  + f)i  ’ 


d?u  _ y2 


dx 2 
d2u 

df 


(x2  + y 2)* 
x 2 

W+lf)r 


du 

dy 

d2,u 

dxdy 

d?u 

dx? 


y 


{f  + y 2)* 
— xy 

{x2  + y2)§ 
— 3 xy2 


d3u  (2r2  — y2)  _ 

dx2dy  ~ y (x2  + y2)?’ 

d3u  _ — 3 yx2 


(x2  + ylY 
dhi  _ x (2  y2  — x 2) 
dxdy2  ~ (a4  + 


dy3  {. x 2 + y2)  t 


d3u  = [ — 3 xy2dx3  + 3 y (2x2  — y2)  dx2dy 
+ dx  (2 y2  — x2)  dxdy2  — 3 yx2dy3\ 


(x2  -f-  y2)% 

2.  u — e(ax+r,y\ 

d?u  — \_a\lx2  + 2 abdxdy  + b2dy2\  e<tc+6» 

= [adx  -f-  bdyY  eax+iy. 

86.  Implicit  Functions  (see  Art.  6). — Thus  far  in  this 
Chapter,  the  methods  which  we  have  given,  although  often 
convenient,  are  not  absolutely  necessary,  as  in  every  case  by 
making  the  proper  substitutions  we  may  obtain  an  explicit 
function  of  x,  and  differentiate  it  by  the  rules  in  Chapter  II. 
But  the  case  of  implicit  functions  which  we  are  now  to 
consider  is  one  in  which  a new  method  is  often  indis- 
pensable. 

Let  / ( x , y)  = 0 be  an  implicit  function  of  two  varia- 

du 

bles,  m which  it  is  required  to  find  If  this  equation 


136 


IMPLICIT  FUNCTIONS. 


can  be  solved  with  respect  to  y,  giving  for  example 
y — (f>  ( x ),  then  the  derivative  of  y with  respect  to  x can  be 
found  by  previous  rules.  But  as  it  is  often  difficult  and 
sometimes  impossible  to  solve  the  given  equation,  it  is 


necessary  to  investigate  a rule  for  finding  without 
solving  the  equation. 


87.  Differentiation  of  an  Implicit  Function. 

Let  f{x,  y)  = 0, 


in  which  y is  an  implicit  function  of  x,  to  find 
Let  f{x,  y)  =.  u. 

Then  u = f (x,  y)  = 0. 

Hence  by  (Art.  82,  Cor.  1),  we  have. 


a 'y e 
(Lx 


ldu\ 
\dx)  ~ 


du,  du  dy 

dx  dy  dx 


But  u is  always  = 0,  and  therefore  its  total  differential 

= 0 ; hence  {^j  — and  therefore, 

du  du  dy  __  n 
dx  ^ dy  dx  ~ ’ 

from  which  we  get, 

du 

dy  _ dx 

dx  ~ du’ 

dy 

Sch. — It  will  be  observed  that  while  = 0,  neither 

For  example. 


iu  du  . . , 

-=j—  nor  is  m general  = 0 
dx  dy 


0 


x2  + y2  — r2 

is  of  the  form  f(x,  y)  = 0.  We  see  that  if  x changes  while 
y remains  constant,  the  function  changes,  and  hence  is  no 


EXAMPLES. 


137 


longer  = 0.  Also,  if  y changes  while  x remains  constant, 
the  function  does  not  remain  = 0.  But  if  when  x changes 
y takes  a corresponding  change  by  virtue  of  its  dependence 
on  x,  the  function  remains  =:  0. 


EXAMPLES. 


1.  y 2 — 2xy  + a2  — 0,  to  find 


dy 


du 

aZ  = “ % ; 


dx 

du 


dx 


dy 


du 

dx 


dy 


= 2 y — 2x. 


therefore,  ~ = T-  = — - 

dx  ■ 


2y 


y 


du 

dy 


2y  — 2x  y — x 


2.  a2y2  + b2x2  — a2b 2 = 0,  to  find 


dJL. 

dx 


du 

dx 


- 2 b2x; 
du 


| = 


„ dy  dx  2b2x  b2x 

therefore,  -j—  — j ^ — — 

clx  du  2a2y 


du 

dy 


a2y 


b 


0) 


Since  y — - V a2  — x2,  from  the  given  equation,  we  may 
solve  this  example  directly  by  previous  methods,  and  obtain 
dy  _ bx 


dx 


aV  a2  — x2> 


(2) 


which  agrees  with  (1)  by  substituting  in  it  the  value  of  y in 
terms  of  x. 

In  this  example  we  can  verify  our  new  rule  by  comparing 
the  result  with  that  obtained  by  previous  rules.  In  more 
complex  examples,  such  as  the  following  one,  we  can  find 
du 

^ only  by  the  new  method. 


138  SECOND  DERIVATIVE  OF  AN  IMPLICIT  FUNCTION 


3.  x?  — ax?y  + bx2y2  — y5  — 0,  to  find  -y-- 

(JbJL 

(1/11 

— = 5x?  — 3 ax2y  + 2 bxy2,} 


therefore, 


~ = _ ax3  + 2bxhj  — 5 y* ; 

dy  5x?  — 3ax2y  + 2 bxy% 

dx  5 yi  — 2 bx2y  + ax? 


4.  ax?  -f  xsy  — ay2  = 0. 


dy  _ 3 ax?  -f  3 x?y 

dx  ~ x?  — 3rn/2 


5.  ?/2  — 2«a;y  + x2  — l?  — 0. 


dy  _ ay  — x 
dx  y — ax 


6.  ys  — 3y  + x ~ 0. 

7.  rc8  + 3axy  -f  ?/3  = 0. 


dy  _ 1 

dr  — 3 (1  — y2) 

d}[  _ _ x?  + ay 
dx  ~ y2  + ax 


88.  To  Find  the  Second  Derivative  of  an  Im 
plicit  Function. 


Let 


We  have 


or 


u =f(x>  y)  = o. 

du 

dy  dx  .. 

S = - Tu  <Art  8?>- 


du  du  dy > 

dx  dy  dx  ’ 

d2y 


dy 


it  is  required  to  find 

(I'll  du 

Differentiating  (2),  remembering  that  , 

tions  of  x and  y,  we  get 


(1) 

(2) 


are  func 


d2u  d2u  dy  id?u  dy  d2u  \dy  du  d?y  _ 
dx2  dy  dx  dx  \dy2  dx  dx  dy)  dx  dy  dx? 


EXAMPLES. 


139 


nr  — , o &u  fy  <ffudif  dud2y_  . . 

dx2  + dxdy  dx  + ^ + dy  dx2  ~ 

dll 

Substituting  the  value  of  ~ from  (1),  and  clearing  of 
fractions,  we  get 
d?u  du2  % 

dx 2 dy2  dx  dy  dx  dy 


cPu  du  du  d2u  du 2 did  d2y  

+ dfdx2  + df  dx*  ~ U’ 


cfiy 

Solving  for  , we  get 


cPy 
dx 2 


ddu  IdiCd 
dx 2 \dy) 


dhi  du  du 
dx  dy  dx  dy 
Idud 

\ dy> 


+ 


d?u  I dud 
dy 2 \ dx) 


(4) 


Sch. — This  equation  is  so  complicated  that  in  practice  it 
is  generally  more  convenient  to  differentiate  the  value  of  the 
first  derivative  immediately  than  to  substitute  in  (4).  The 
third  and  higher  derivatives  may  be  obtained  in  a similar 
manner,  but  their  forms  are  very  complicated. 

Equation  (2)  is  frequently  called  the  first  derived  equation 
or  the  differential  equation  of  the  first  order  j and  equation 
(3)  is  called  the  second  derived  equation  or  the  differential 
equation  of  the  second  order. 


EXAMPLES. 


1.  y 2 — 2 xy  + a2  — 0,  to  find  ^ and  ^ 


dx 2 
cPu 
dx 2 


= o ; 


cPu 
dx  dy 


= -2; 
dy  _ 


d2u 

dy2 


= 2. 


Therefore,  by  (1),  fx~y_x, 

and  bv  ^ - ~ 16y  (y  ~ s)  + 8.V3  _ V (d  ~ 

J { h dx2  ~ (2 y - 2a:)3  “ {y  - x)3  ' 


140 


CHANGE  OF  INDEPENDENT  VARIABLE. 


dx 2 


2.  y 2 — 2axy  + x2  — V2  = 0,  to  find  (See  Sch.) 

ay  — x 

dx  ~ y — ax 

(*/  — az)2 

_ (y  — gar)  (a2y  — y)  — {ay  — x)  (a2x  - x) 

(. V — ax)3 

(by  substituting  the  value  of  ('j' ) 

(a2  — 1)  (y2  — 2axy  x2)  _V2  (a2  — 1) 

(; y — ax )3  ~ (y  — ax)3  ’ 

3.  2s  + 3 axy  + y3  = 0,  to  find  ^ and 

ClCC* 

Differentiating,  we  have 

(x2  + ay)  dx  + (y2  -f  ax)  dy  = 0 ^ 

dy  a?  -f  ay 

dx  y2  + ax 

(a x + (y2  + ax)  ~(?y%  + a ) (**+«$ 

(y2  + ax)2 

(See  Price’s  Calculus,  Vol.  I,  p.  142.) 

2 


cPy 
dx2  ~ 


2a3xy 


{y2  + ax)3 
4.  x2  -f-  y2  — a2  = 0. 


dy x _ <$y  _ a 

dx  y ’ dx 2 y3 


89.  Change  of  the  Independent  Variable. — Thus 

far  we  have  employed  the  derivatives  ~2,  etc.,  upon 

the  hypothesis  that  x was  the  independent  variable  and  y 
the  function.  But  in  the  discussion  of  expressions  contain- 


VALUES  OF  DIFFERENTIAL  COEFFICIENTS.  141 


ing  the  successive  differentials  and  derivatives  of  a function 
with  respect  to  x,  it  is  frequently  desirable  to  change  the 
expression  into  its  equivalent  when  y is  made  the  independ- 
ent variable  and  x the  function  ; or  to  introduce  some  other 
variable  of  which  both  y and  x are  functions,  and  make  it 
the  independent  variable. 


90.  To  Find  the  Values  of  f~ , etc., 

5 dx  dx2  dx3 

when  neither  x nor  y is  Equicrescent.  (Art.  55.) 


The  value  of  the  first  derivative,  C~,  will  be  the  same 

dx 

whether  x or  y,  or  neither,  is  considered  equicrescent. 

Cp’V 

The  value  of  the  second  derivative,  ~ , was  obtained  in 

dx1 

dv 

Art.  56  by  differentiating  as  a fraction  with  a constant 
denominator  and  dividing  by  dx. 


If  we  now  consider  that  neither  x nor  y is  equicrescent, 
and  hence  that  both  dx  and  dy  are  variables,  and  differen- 
tiate we  have 
dx 

d?y  d ldy\  d2y  dx  — cPx  dy  . 

d = hw  = <l> 


which  is  therefore  the  value  of  the  second  derivative  when 
neither  variable  is  equicrescent. 

Similarly, 

d3y  d lcPy\ 

dx2  dx\dxV 


(d3y  dx  — cPx  dy)  dx  — 3 ((Py  dx  — d?x  dy)  tPx 
dx5 


(2) 


which  is  the  value  of  the  third  derivative  when  neither 
variable  is  equicrescent,  and  so  on  for  any  other  derivative. 


142 


EXAMPLES. 


Cor. — If  x is  equicrescent,  these  equations  are  identical. 
If  y is  equicrescent,  d2y  = d3y  — 0,  and  (1)  becomes 


cPy  d?x  dy 

dx2  ~ c lx?  ’ 

and  (2)  becomes 

d3y  _ 3 (d2x)2  dy  — d3x  dy  dx 
dx3  “ !hF~ 


(3) 

(4) 


which  are  the  values  of  the  second  and  third  derivatives 
when  y is  equicrescent. 


Sch.  1. — Hence,  if  we  wish  to  change  an  expression  when 
x is  equicrescent  into  its  equivalent  where  neither  x nor  y is 

cfiy  dhi 

equicrescent,  we  must  replace  ^ , etc.,  by  their  com- 


plete values  in  (1),  (2),  etc.;  but  if  we  want  an  equivalent 
expression  in  which  y is  equicrescent,  we  must  replace 

(~2,  , etc.,  by  their  values  in  (3),  (4),  etc. 


Sch.  2. — If  we  wish  to  change  an  expression  in  which  x 
is  equicrescent  into  its  equivalent,  and  have  the  result  in 
terms  of  a new  independent  variable  t,  of  which  a;  is  a 

(Py  (Py 

function,  we  must  replace  f , , etc.,  by  their  complete 


values  in  (1),  (2),  etc.,  and  then  substitute  in  the  resulting 
expression,  in  which  neither  x nor  y is  equicrescent,  the 
values  of  x,  dx,  d2x,  etc.,  in  terms  of  the  new  equicrescent 
variable. 


EXAMPLES 

1.  Transform  x ^ + ('y  ) — ^ = 0,  in  which  x is 

equicrescent,  into  its  equivalents,  (1)  when  neither  x nor  y 
is  equicrescent,  (2)  when  y is  equicrescent. 


EXAMPLES. 


143 


(1.)  Replace  y-|  by  its  value  in  (1),  and  multiply  by  dx% 
and  we  have 

x ( d2y  dx  — dzx  dy)  + dy3  — dy  dx?  = 0. 

(2.)  Put  d?y  = 0,  divide  by  dy3  to  have  the  differential 
*)f  the  independent  variable  in  its  proper  position,  the  de- 
nominator, and  change  signs,  and  we  have 


dy 2 


/dx\2 

[dy 


2.  Transform  yy  — — y + T" 
dx 2 1 — x?  dx  1 


y 


0,  in  which  x 


is  equicrescent,  into  its  equivalent  when  6 is  equicrescent, 
having  given  x — cos  9. 

Replacing  C~  by  its  complete  value  in  (1),  the  given 

equation  becomes 

d?y  dx  — d?x  dy 
dx? 


dy 


+ 


y _ 


= o. 


1 — x 2 dx  1 — x2 

x — cos  0 ; 

dx  = — sin  9 d6  and  cPx  = — co»  6 d9\ 
1 — x2  = sin2  9. 

Substituting,  we  have 


— cPy  sin  9 dd  -f  cos  9 dO 2 dy  cos  9 dy 

— sin3  6 dd 3 sin2  9 sin  9 d9 


+ mrVTi  n 7 a -H  " • ^ 


JL  - 

sin20 


d?y 
dd 2 


+ y — 0.  (See  Price’s  Calculus,  Yol.  I,  p.  126.) 


dx 2 


Transform  + y = 0,  in  which  x is 

oc  ctoc 


eqm- 


crescent,  into  its  equivalent,  (i)  when  neither  y nor  9 is 
equicrescent ; (2)  when  9 is  equicrescent ; (3)  when  y is 
equicrescent,  having  given  x?  — 4 9. 


144 


EXAMPLES. 


Replacing  in  this  equation  the  complete  value  of  it 
becomes 

iP-y  dx  — d?x  dy  1 dy  _ 
cfo3  x dx  y 


x — 2 (6)s  ; 


dx  = d~?dO. 


^ = 5 + T* 

20*  6 i 

Substituting,  we  have 

(2.)  y dd3  + dy  dd2  -f  0 cPy  dd  — d dy  d26  — 0. 

(2.)  When  6 is  equicrescent,  d2d  — 0;  therefore  (1)  be- 


comes 


or 


y dd 3 + dy  dd 2 -f  0 d2y  dd  = 0, 

e^l  + <*l  + » = o. 

dd2^  dd^  y 


(8.)  When  y is  equicrescent,  cPy  = 0 ; therefore  (1)  be- 
comes 


e 


(Pd  tdd\ 2 
df 


dy 

1 + 


4.  Transform  R — 


(dd\2  /ddy  _ A 
\<r/y/  ^ W#/ 

[ 


^7 

dx2 J 


efa2 


into  its  equivalent,  (2)  in 


the  most  general  form ; (2)  when  d is  equicrescent ; (5) 
when  r is  equicrescent,  having  given  x = r cos  d,  and 
y — r sin  d. 

The  complete  value  of  R is 

T?  - _ (^  + <%»)* 

(Pxdy  — cPydx 

dx  = dr  cos  d — r sin  d d d, 


dy  — dr  sin  d + r cos  d d Q. 


EXAMPLES. 


145 


cPx  = cos  6 cPr  — 2 sin  6 drdO  — r cos  6 d(P 

— r sin  6 cPO, 

d?y  — sin  0 d2r  + 2 cos  6 drdO  — r sin  6 dd 2 

+ r cos  6 cPO  ; 

(dx)2  + ( dy )2  = dr 2 + r2dQ2, 

cPxdy  — d2ydx  = rcPrdO  — 2 dr2dd  — r2dd3  — rdrdPQ. 

m . p [dr2  + rW]^ 

' *'  ' ' 1 — rcPrdO  q-  2dr2dQ  + r^dd3  + rdrd20} 


(See  Serret’s  Calcul  Differential  et  Integral,  p.  94.^ 

5.  Transform 

(i dy 2 + dx2)%  -f  adxdPy  = 0, 

in  which  x is  equicrescent,  into  its  equivalents,  (1)  when 
neither  x nor  y is  equicrescent,  (2)  when  y is  equicrescent. 

(1.)  (dy2  + dx2)§  + a (d2ydx  — cPxdy)  — 0 ; 


91.  General  Case  of  Transformation  for  Two  Inde- 
pendent Variables. — Let  u be  a function  of  the  inde- 
pendent variables,  say  u — f (x,  y) ; and  suppose  x and  y 
functions  of  two  new  independent  variables  r and  0, 
so  that, 


7 


146 


EXAMPLES. 


x — <f>  (r,  d)  and  y = ip  (r,  6) ; (1) 


then  u may  be  regarded  as  a function  of  r and  d,  through 

du  du 

x and  y.  It  is  required  to  find  the  values  of  -j-  and  — in 

(XX  (JL  if 

terms  of  derivatives  of  u,  taken  with  respect  to  the  new 
variables  r and  d. 


Since  u is  a function  of  r through  x and  y,  we  have 
(Art.  82), 


And  similarly 


du  du  dx  du  dy 

dr  dx  dr  dy  dr 


(2) 


du 

dd 


du  dx 
dx  dd 


du  dy  % 
dy  dd’ 


where  the  values  of  -7- , -p- 


dx 

dr 


dy_ 

dr ! 


dx 

dd’ 


dy 

dd’ 


can  be  found  from 


(3) 

(1). 


Whenever  equations  (1)  can  he  solved  for  r and  6 sepa- 
rately, we  can  find  by  direct  differentiation  the  values  of 

dr  dr  dd  dd  . , ... 

— , — , — , — , and  hence  by  substituting  in 

CLjC  till  (IX  (tlj 


du  du  dr 

dx  dr  dx 


du  dd 
+ Id  dx’ 


and 


du  du  dr  du  dd 

^ = S^  + 55*,<Al-t82>’ 


we  can  obtain  the  values  of  ^ and  —• 

dx  dy 

When  this  process  is  not  practicable,  we  can  obtain  their 
values  by  solving  (2)  and  (3)  directly,  as  follows : 

flqr  fly 

Multiply  (2)  by  ^ and  (3)  by  and  subtract ; then 

multiply  (2)  by  ~ and  (3)  by  ^ and  subtract.  We  shall 
then  have  two  equations,  from  which  we  obtain, 


EXAMPLES. 


147 


du 

dy 

du 

dy 

du 

dr 

dd 

dd 

dr 

dx 

~ dx 

dy 

dy 

dx 

dr 

dd 

dr 

dd 

du 

dx 

du 

dx 

du 

dd 

dr 

dr 

dd 

dy  ~ 

dx 

dy 

dy 

dx 

dr 

dd 

dr 

dd 

(4) 

(5) 


The  values  of  etc.,  can  be  obtained  from  these, 

but  the  general  formulae  are  too  complicated  to  be  of  much 
practical  use.  (See  Gregory’s  Examples,  p.  35.) 


Cor. — If  x = r cos  d and  y = r sin  d,  (4)  and  (5) 
become 


du  ^ du  ton  0 du  _ du 

dx  ~ C0S  dr  r dd  ’ dy 


cos  6 du  . . 

m + sm  6 

r dd 


du 

dr 


1. 


2. 


3. 


4. 


EXAMPLES. 

u = to  find  du  (Art  81). 

x — y 

_ 2 { rdy  - ydx) 
dU~  (x  - yf 

u = sin  ax  + sin  by  + tan-1 

du  = a cos  axdx  + b cos  bydy  + 


u — sm  1 -• 

y 

_ yd'x  — xdy 
y Vy2  — x2 
u = sin  ( x + y). 

du  = cos  {x  + y)  (dx  + dy). 


zdy  — ydz 

y%  h z2 


148 


EXAMPLES. 


6.  U = 
6.  u — 

? U — 

8.  u = 
(Art.  82). 

9.  u = 

10.  u 

11.  it 

12.  If 

cPu 

dxdy 


z4  — a4 


7 _ x (z2  — a2)  (2ydx  + xdij)  — 2x2t/zdz 
du  - ’ (z2  - 0 " 


log 


lx  + V%2  — y2\ 
\x  — V x2  — v‘ 


du  = 2 


y<?a:  — xdy 
y x2  — y2 

/du\ 


cot  xy  to  find  (Art.  82,  Cor.  1). 

(£)  = -a'cWa'S+logit)- 

sin  (z/2  — z),  and  y — log  a-,  to  find 


c7;A  _2{y  — x2)  cos  (y2  — g) 
^a;/  _ £ 


log  tan  — 

° y 


(du\ 

\dx)  “ ” 


dy 

y-xtx 


y2  sin  - cos  - 

9 y y 


log  ( x — a -j-  V -f2  — 2aa;). 

/r/w\  1 

\5x/  _ aA-2  _ 2, 


V a;2  — 2aa: 
w = a^z4  + erz/2z3  -f-  x2y2z2,  show  that 

= Qezyz2  -f  8 yz.  (Art.  83.) 

Xy , show  that 

15  xy 


diu 

dx2dydz 
u = tan-1 


Vi  + x1  + y2' 

1 d*u 


(1  + x*  + y2)? 


*’  dx2dy2  (i+a^  + y*)* 


EXAMPLES. 


149 


13.  u = x?y2  + y3x?  to  find  d2u  (Art.  85). 

= Gxy2dx 2 + 6x2ydxdi / + 2 y3dx2  + 6 xy2dxdy 
+ 6x2ydxdy  + 2 x?dy2  + 6 xifdxdy  + dxh/dy2 
— ( 6xy2  + 2 y3)  dx2  + 12  ( x2y  + y2x)  dxdy  + ( 6 x2y  + 2a^)  dy2. 


14.  xy  + yx—  a = 0,  to  fi 

15.  xy  — y1  = 0. 

16.  y2  (2 a — x)  — x3  = 0. 

17.  x?  — daxy  + y3  = 0. 

18.  yenv  — axm  — 0. 

19.  a?  + Vsec  {xy)  — 0. 

dy  _ y V sec  {xy)  ta 

dx  x \/ sec  ( xy ) tan 


ld  % (Art- sz) 

dy  _ _ yxy~x  + y log  y 

dr  xy1'1  + xu  log  x 

dy  _ y2  — xy  log  y 
da  a:2  — xy  log  a; 

d?/  3.+  + t/2 

da;  — 2 y (2a  — a;) 

ah/ ay  — a;2 

dx  y2  — ax 

dy my 

dx  ~ ' x (1  + ny) 

n [xy)  + 2alVyxv~1  log  a 
{xy)  + 2azVxy  log  a log  x 


20.  xi  + 2 ax2y 


dy  (py 


ay3  = 0,  to  find  ~ and 


dx2 


(Art.  88.) 


dy  4ar3  + 4 axy 

dx  2aa;2  — 3 ay2 


= — [(12a;2  + 4 ay)  {2ax2  - day2)2 

— 8aa;  (4a-3  + kaxy)  (2 ax2 — day2)  — (4a^  + 4 axy)2  day 

— (2 ax2  — day2)3  = what  ? 


Show  that 


dy 

dx 


= 0 or  ± \/2,  when  x = 0 and  y = 0- 


150 


EXAMPLES. 


21,  Change  the  independent  variable  from  x to  t in 

(1  — a,-2)  — x = 0,  when  x = cos  t. 

Ans  - f 
Am.  dp  - 0 

22.  Change  the  independent  variable  from  x to  8 m 


d2y  2x  dy 

~JPb  “r  i , Vo  T YZ  + 


y 


dx2  1 + xz  dx  (1  + z2)2 


0,  when  x = tan  8. 

. cPy  n 

Ans-  + y = °- 


23.  Change  the  independent  variable  from  x to  r,  and 
eliminate  x,  y,  dx  and  dy,  between 

t = -!l!L — x = r cos  8,  and  y = r sin  0. 
ydy  xdx 

Am.  t = "'l 
dr 

24.  Change  the  independent  variable  from  x to  z in 

P —.  x2  — 2x  + y when  x = d and  y = er(v  4 2). 


dx2  ‘ dx  J dx’ 

(]%i)  fly 

Am.  P = ez  -g  + (v  •+  1 ) ^ +■  v {v  + 2) 


} 


CHAPTER  VIII. 


MAXIMA  AND  MINIMA  OF  ^UNCTIONS  OF  A 
SINGLE  VARIABLE. 


92.  Definition  of  a Maximum  and  a Minimum. — if, 

while  the  independent  variable  increases  continuously,  a 
function  dependent  on  it  increases  up  to  a certain  value, 
and  then  decreases,  the  value  at  the  end  of  the  increase  is 
called  a maximum  value  of  the  function. 

"i  while  the  independent  variable  increases,  the  function 
decreases  to  a certain  value  and  then  increases,  the  value  at 
the  end  of  the  decrease  is  called  a minimum  value  of  the 
function.  Hence,  a maximum  value  of  a function  of 
a single  variable  is  a value  which  is  greater  than  the 
immediately  preceding  and  succeeding  values,  and  a 
minimum  value  is  less  than  the  immediately  pre- 
ceding and  succeeding  values. 

For  example,  sin  0 increases  as  0 increases  till  the  latter 
reaches  90°,  after  which  sin  0 decreases  as  0 increases  ; 
that  is,  sin  0 is  a maximum  when  0 is  90°,  since  it  is 
greater  than  the  immediately  preceding  and  succeeding 
values.  Also,  cosec  0 decreases  as  0 increases  till  the  latter 
reaches  90°,  after  which  cosec  0 increases  as  0 increases  ; 
that  is,  cosec  0 is  a minimum  when  0 is  90°,  since  it  is  less 
than  the  immediately  preceding  and  succeeding  values. 


93.  Condition  for  a Maximum  or  Minimum. — If  y 

be  any  function  of  x,  and  y be  increasing  as  x increases, 
the  differential  of  the  function  is  positive  (Art.  12),  and 

dy 


hence  the  first  derivative  — will  be  positive. 


If  the  func- 
tion be  decreasinci  as  x increases,  the  differential  of  the 


152 


GEOMETRIC  ILLUSTRATION. 


function  is  negative,  and  hence  the  first  derivative  ^ will 
be  negative. 

Therefore,  since  at  a maximum  value  the  function 
changes  from  increasing  to  decreasing,  the  first  derivative 
must  change  its  sign  from  plus  to  minus ; as  the  variable 
increases.  And  since,  at  a minimum  value,  the  function 
changes  from  decreasing  to  increasing,  the  first  derivative 
must  change  its  sign  from  minus  to  plus.  But  as  a function 
which  is  continuous*  can  change  its  sign  only  by  passing 
through  0 or  go  , it  follows  that  the  only  values  of  the 
variable  corresponding  to  a maximum  or  a minimum 
value  of  the  function,  are  those  which  make  the  first 
derivative  O or  oc . 


94.  Geometric  Illustra- 
tion. — This  result  is  also 
evident  from  geometric  con- 
siderations ; for,  let  y — f(x ) 
be  the  equation  of  the  curve 
AB.  At  the  points  P,  P',  P", 

Piy,  the  tangents  to  the  curve 
are  parallel  to  the  axis  of  x, 
and  therefore  at  each  of  these  points  the  first  derivative 
/'  ( x ) = 0,  by  Art.  56a. 

We  see  that  as  x is  increasing  and  y approaching  a 
maximum  value,  as  PM,  the  tangent  to  the  curve  makes 
an  acute  angle  with  the  axis  of  x ; hence,  approaching  F 

from  the  left  -f-  is  + . At  P the  tangent  becomes  parallel 
dx 

to  the  axis  of  x ; hence,  ~ = 0.  Immediately  after  pass- 
clx 

ing  P the  tangent  makes  an  obtuse  angle  with  the  axit 
of  x ; hence,  — is  — . 

CbX 


Y 


* In  this  discussion  the  function  is  to  he  resarde'I  as  conMnuoiui. 


CRITERION  OF  MAXIMA  AND  MINIMA. 


153 


Also  in  approaching  a minimum  value,  as  P'M',  from  the 
left,  we  see  that  the  tangent  makes  an  obtuse  angle  with 

the  axis  of  x,  and  hence  — is  — . At  the  point  P',  ~ = 0. 

clx 

Aiter  passing  P',  the  angle  is  acute  and  ~ is  -f-. 

ax 

dv 

In  passing  P'",  ^ changes  sign  by  passing  through  oo , 

P"'M"'  is  a minimum  ordinate.  In  approaching  it  from  the 
left  the  tangent  makes  an  obtuse  angle  with  the  axis  of  x, 

and  hence  — is  — . At  P"'  the  tangent  is  perpendicular 


to  the  axis  of  x,  and  — cc . 

ax 


After  passing  P'"M'",  the 


angle  is  acute  and  ~ is  +. 

& dx 

While  the  first  derivative  can  change  its  sign  from  + to 
— or  from  — to  + only  by  passing  through  0 or  oo , it 
does  not  follow  that  because  it  is  0 or  oo,  it  therefore 
necessarily  changes  its  sign.  The  first  derivative  as  the 
variable  increases  may  be  +,  then  0,  and  then  +,  or  it  may 
be  — , then  0,  and  then  — . This  is  evident  from  Fig.  9, 
where,  at  the  point  D,  the  tangent  is  parallel  to  the  axis  of 
dy 

x,  and  is  0,  although  just  before  and  just  after  it  is  — . 


Hence  the  values  of  the  variable  which  make  — 0 or  qo  , 

dx 

are  simply  critical * values,  i.  e.,  values  to  be  examined. 

As  a maximum  value  is  merely  a value  greater  than  that 
which  immediately  precedes  and  follows  it,  a function  may 
have  several  maximum  values,  and  for  a like  reason  it  may 
have  several  minimum  values.  Also,  a maximum  value 
may  be  equal  to  or  even  less  than  a minimum  value  of  the 
same  function.  For  example,  in  Fig.  9,  the  minimum  P'M' 
is  greater  than  the  maximum  PivMiv. 


See  Price’s  Cal.,  Vol.  I,  p.  337, 


154  CONDITION  GIVEN  BY  TAYLOR'S  THEOREM. 

95.  Method  of  Discriminating  between  Maxima 
and  Minima. — Since  the  first  derivative  at  a maximum 
state  is  0,  and  at  the  immediately  succeeding  state  is  — , it 
follows  that  the  second  derivative,  which  is  the  difference 
between  two  consecutive  first  derivatives,*  is  — at  a maxi- 
mum. Also,  since  the  first  derivative  at  a minimum  state 
is  0,  and  at  the  immediately  succeeding  state  is  + , it  fob 
lows  that  the  second  derivative  is  + at  a minimum.  There- 
fore, for  critical  values  of  the  variable,  a function  is  at  a 
maximum  or  a minimum  state  according  as  its 
second  derivative  at  that  state  is  — or  + . 

96.  Condition  for  a Maximum  or  Minimum  given 
by  Taylor’s  Theorem. — Let  u — fix)  be  any  continuous 
function  of  one  variable ; and  let  a be  a value  of  x corre- 
sponding to  a maximum  or  a minimum  value  of  fix). 
Then  if  a takes  a small  increment  and  a small  decrement 
each  equal  to  h,  in  the  case  of  a maximum  we  must  have, 
for  small  values  of  h, 

f{a)>f(a  + h)  and  f(a)>f(a  — h)-, 
and  for  a minimum, 

f{")<f{a  + li)  and  f {a)  < f {a  — h). 

Therefore,  in  either  case, 

f[a  + h)  — f(a)  and  f(a  — U)  — f{a) 
have  both  the  same  sign. 

By  Taylor’s  Theorem,  Art.  66,  Eq.  7,  and  transposing, 
we  have 

f(a  + h)—  f(a)~  f (a)  h +/"(«)  | + etc.;  (1) 

Id 

/(«  — li)  —/(a)  = -/'  (a)  h +f"  (a)  - — etc.  (2) 


Remembering  that  the  first  value  is  always  to  be  subtracted  from  the  second. 


FINDING  MAXIMA  AND  MINIMA  VALUES. 


155 


Now  if  h be  taken  infinitely  small,  the  first  term  in  the 
second  member  of  each  of  the  equations  (1)  and  (2)  will  be 
greater  than  the  sum  of  all  the  rest,  and  the  sign  of  the 
second  member  of  each  will  be  the  same  as  that  of  its  first 
term,  and  hence  / (a  + h)  — f (ci)  and  f (a  — h) — f{a) 
cannot  have  the  same  sign  unless  the  first  term  of  (1)  and 
(2)  disappears,  which,  since  li  is  not  0,  requires  that 
/'( a)  - 0. 

Hence,  the  values  of  oc  which  make  f{x)  a maxi- 
mum or  a minimum  are  in  general  roots  of  the  equa- 
tion, f'  ( x ) = 0. 

Also,  when  /'  (a)  = 0,  the  second  members  of  (1)  and 
(2),  for  small  values  of  h,  have  the  same  sign  as  /"  (a); 
that  is,  the  first  members  of  (1)  and  (2)  are  both  positive 
when  f"  (a)  is  positive,  and  negative  when  f"  (a)  is  nega- 
tive. Therefore,  f{ a)  is  a maximum  or  a minimum 
according  as  f"  («)  is  negative  or  positive. 

If,  however,  f"  {a)  vanish  along  with  f (a),  the  signs  of 
the  second  members  of  (1)  and  (2)  will  be  the  same  as 
/"'  (a),  and  since  /"'  (a)  has  opposite  signs,  it  follows  that 
in  this  case  f(ci)  is  neither  a maximum  nor  a mini- 
mum unless  (a)  also  vanish.  But  if  /"'  («)  = 0, 
then  f(a)  is  a maximum  when  f iv  (a)  is  negative,  and  a 
minimum  when  f"  (a)  is  positive,  and  so  on.  If  the  first 
derivative  which  does  not  vanish  is  of  an  odd  order,  f(a)  is 
neither  a maximum  nor  a minimum  ; if  of  an  even  order, 
f(a)  is  a maximum  or  a minimum,  according  as  the  sign 
of  the  derivative  which  does  not  vanish  is  negative  or  posi- 
tive. 

97.  Method  of  Finding  Maxima  and  Minima 
Values. — Hence,  as  the  result  of  the  preceding  investiga- 
tion we  have  the  following  rule  for  finding  the  maximum 
or  minimum  values  of  a given  function,  f(x). 

Find  its  first  derivative,  f'  {pc)  put  it  equal  to  0, 
and  solve  the  equation  thus  formed,  f ( x ) = 0.  Sub- 


156  MAXIMA  AND  MINIMA  VALUES  ALTERNATE. 

stitute  the  values  of  x thus  found  for  x in  the  second 
derivative,  f"  (sc).  Each  value  of  x ivhich  makes 
the  second  derivative  negative  will,  when  substituted 
in  the  function  f (x)  make  it  a maximum ; and  each 
value  ivhich  makes  the  second  derivative  positive  will 
make  the  function  a minimum.  If  either  value  of 
x reduces  the  second  derivative  to  0,  substitute  in  the 
third,  fourth,  etc.,  until  a derivative  is  found  which 
does  not  reduce  to  0.  If  this  be  of  an  odd  order,  the 
value  of  x will  not  make  the  function  a maximum 
or  minimum ; but  if  it  be  of  an  even  order  and  nega- 
tive, the  function  will  be  a maximum ; if  positive,  a 
minimum. 

Second  Rule. — It  is  sometimes  more  convenient  to 
ascertain  whether  a root  a of/'(.r)  = 0 corresponds  to  a 
maximum  or  a minimum  value  of  the  function  by  substi- 
tuting for  x,  in  f ( x ),  a — h and  a + h,  where  h is  infini- 
tesimal. If  the  first  result  is  + and  the  second  is  — , 
a corresponds  to  a maximum ; if  the  first  result  is 
— and  the  second  is  +,  it  corresponds  to  a minimum. 
If  both  results  have  the  same  sign,  it  corresponds  to 
neither  a maximum  nor  a minimum.  (See  Arts.  93,  94.) 

98.  Maxima  and  Minima  Values  occur  alternately. 

— Suppose  that  f (x)  is  a maximum  when  x — a,  and  also 
when  x = b,  where  b > a ; then,  in  passing  from  a to  b, 
when  x — a + h (where  h is  very  small),  the  function  is 
decreasing,  and  when  x — b — li,  it  is  increasing;  but  in 
passing  from  a decreasing  to  an  increasing  state,  it  must 
pass  through  a minimum  value  ; hence,  between  two  maxi- 
ma one  minimum  at  least  must  exist. 

In  the  same  way,  it  may  be  shown  that  between  two 
minima  one  maximum  must  exist. 

This  is  also  evident  from  geometric  considerations,  for  in 
Fig.  9 we  see  that  there  is  a maximum  value  at  P,  a mini- 
mum at  P',  a maximum  at  P",  a minimum  at  P”',  and  so  on. 


APPLICATIONS  OF  AXIOMATIC  PRINCIPLES.  157 


99.  The  Investigation  of  Maxima  and  Minima  is 
often  facilitated  by  the  following  Axiomatic  Prin- 
ciples : 

1.  If  u be  a maximum  or  minimum  for  any  value  of  x, 
and  a be  a positive  constant,  au  is  also  a maximum  or  mini- 
mum for  the  same  value  of  x.  Hence,  before  applying  the 
rule,  a constant  factor  or  divisor  may  be  omitted. 

2.  If  any  value  of  x makes  u a maximum  or  minimum, 
it  will  make  any  positive  power  of  u a maximum  or  mini- 
mum, unless  u be  negative,  in  which  case  an  even  power  of 
a minimum  is  a maximum,  and  an  even  power  of  a maxi- 
mum is  a minimum.  Hence,  the  function  may  be  raised 
to  any  -power ; or,  if  under  a radical,  the  radical  may 
be  omitted. 

3.  Whenever  u is  a maximum  ora  minimum,  log  u is  a 
maximum  or  minimum  for  the  same  value  of  x.  Hence, 
to  examine  the  logarithm  of  a function  ice  have  only 
to  examine  the  function  itself.  When  the  function  con- 
sists of  products  or  quotients  of  roots  and  powers,  its  exam- 
ination is  often  facilitated  by  passing  to  logarithms,  as  the 
differentiation  is  made  easier. 

4.  When  a function  is  a maximum  or  a minimum,  its 
reciprocal  is  at  the  same  time  a minimum  or  a maximum ; 
this  principle  is  of  frequent  use  in  maxima  and  minima. 

5.  If  u is  a maximum  or  minimum,  u ± c is  a maximum 
or  minimum.  Hence,  a constant  connected  by  -f-  or  — 
may  be  omitted. 

Other  transformations  are  sometimes  useful,  but  as  they 
depend  upon  particular  forms  which  but  rarely  occur,  they 
may  be  left  to  the  ingenuity  of  the  student  who  wishes  to 
simplify  the  solution  of  the  proposed  problem. 

It  is  not  admissible  to  assume  x = oo  in  searching;  for 
maxima  and  minima,  for  in  that  case  x cannot  have  a suc- 
ceeding value. 


158 


EXAMPLES. 


EXAMPLES. 


1.  Find  the  values  of  x which  will  make  the  function 
u — 6x  -f-  3a:2  — 4a:3  a maximum  or  minimum,  and  the  cor- 
responding values  of  the  function  u. 

du 

Here  — = 6 + 6a:  — 12a:2. 

ax 

How  whatever  values  of  x make  u a maximum  or  mini- 
clu 

mum,  will  make  — = 0 (Art.  97)  ; therefore, 


6 + 6a;  — 12a;2  = 0,  or  a:2  — \x  — \ ; 

•••  x = i ± f = + 1 or  — 

Hence,  if  u have  maximum  or  minimum  values,  they 
must  occur  when  x = 1 or  — 

To  ascertain  whether  these  values  are  maxima  or  minima, 
we  form  the  second  derivative  of  u;  thus, 


d?u 
clx 2 


6 — 24a;. 


When  x = 1, 


cPu 
clx 2 


= — 18,  which  corresponds  to  a maxi- 


mum value  of  u. 

When  x — — -j~2  = + 18,  which  corresponds  to  a 

minimum  value  of  u. 

Substituting  these  values  of  x in  the  given  function,  we 
have 

When  x = 1,  u = 6 + 3 — 4 = 5,  a maximum. 

When  x——\,  u = — 3 + f + \ — — J,  a minimum 


2.  Find  the  maxima  and  minima  values  of  u in 


u = xt  — 8a^  4-  22a^  — 24a;  + 12. 


^ = 4at>  — 24a;2  + 44a;  — 24  = 0, 
ax 


EXAMPLES. 


159 


or,  z3  — 6a;2  + 11a;  — 6 = 0. 

By  trial,  x = 1 is  found  to  be  a root  of  this  equation ; 
therefore,  by  dividing  the  first  member  of  this  equation  by 
x — 1,  vve  find  for  the  depressed  equation, 

x2  — ox  + 6 = 0;  x = 2 or  3. 

Hence  the  critical  values  are  x — 1,  x — 2,  and  x = 3. 

^ — 12a;2  — 48a;  + 44  = +8,  when  x = 1. 

= — 4,  when  x — 2. 

— + 8,  when  x = 3. 

Therefore  we  have, 

when  x = 1,  m = 3,  a minimum ; 

when  x — 2,  u = 4,  a maximum ; 

when  x — 3,  u = 3,  a minimum. 

3.  Find  the  maxima  and  minima  values  of  u in 
u — (x  — l)4  ( x + 2)3. 

^ = 4 (x  — l)3  (x  + 2f  + 3 (a;  + 2)2(a;  - l)4 
= (*  - l)3  (x  + 2) 2 [4  (x  + 2)  + 3 (x  - 1)], 

nil 

or  ^ = (x  — l)3  ( x + 2)2  (7a;  + 5)  = 0 ; (1) 

(x  — 1)  --  0,  (x  + 2)  = 0,  (7a;  + 5)  = 0. 
x — 1,  x = — 2,  and  x—  — as  the  critical  values 

of  x. 

In  this  case,  it  will  be  easier  to  test  the  critical  values  by 

clw 

the  second  rule  of  Art.  97 ; that  is,  to  see  whether  -T- 

dx 

changes  sign  or  not  in  passing  through  x — 1,  — 2,  and 
— 4 in  succession. 


160 


EXAMPLES. 


It  we  substitute  in  the  second  member  of  (1),  (1  — h] 
and  (1  4 h)  ior  x,  where  h is  infinitesimal,  we  get 

Tx  = (l-/‘-l)*(l-/<  + 2)![7(l-*)+  ?! 

= _ h3  (3  - h)2  (12  - 7 h)  = 

and  ^ = (1+  h — 1)3(1  + h + 2)2 [7  (1  + h)  4-  5] 


= h3{ 3 + h)2  (12  + 7 h)  = + 

chi 

Therefore,  as  ^ changes  sign  from  — to  + at  x = 1 
the  function  u at  this  point  is  a minimum. 
du 

When  x = — 2,  ^ does  not  change  sign ; u has  no 
maximum  or  minimum  at  this  point 
d ic 

When  x — — — changes  sign  from  + to  — ; u, 

at  this  point,  is  a maximum. 

Hence,  when  x — 1,  u — 0,  a minimum. 

. _ 124.93 

when  x = — u — — , a maximum. 


It  is  usually  easy  to  see  from  inspection  whether  — 

changes  sign  in  passing  through  a critical  value  of  x>  with- 
out actually  making  the  substitution. 


4.  Examine  u = b + (x  — a)3  for  maxima  and  minima. 
du 


dx 


= 3 (x  - a)J  = 0 ; 


(Pit 


x = a,  and  u 


Since  x — a makes  ^ = 0,  we  must  examine  it  by  the 

second  rule  of  Art.  97,  and  see  whether  -7-  changes  sign  at 

dx 

* = a. 


EXAMPLES. 


161 


~ = 3 (a  — h — a)2  = 3h2  is  the  value  of  ^ immediately 
preceding  x = a. 

~ = 3 (a  + /t  — a)2  = 3 h2  is  the  value  of  ^ immediately 

succeeding  x = a. 

du 

Therefore,  as  does  not  change  sign  at  x = a,  u = i 
is  neither  a maximum  nor  a minimum. 


5.  Examine  u — b + (x  — «)4  for  maxima  and  minima. 
^ = 4 [x  — «)3  = 0 ; x - : a and  u = b. 


dii 

It  is  easy  to  see  that  changes  sign  from  — to  f at 
r = a ; x — a gives  u — b,  a minimum. 

(cc  — }— 

6.  Examine  u = ) ^ for  maxima  and  minima. 

( x — 3)‘ 

du  (x  + 2)2(x  — 13) 

Tx  = = 0 or 

.*.  x — — 2,  13,  or  3. 


We  see  that  when  x — — 2,  does  not  change  sign  j 
.*.  no  maximum  or  minimum ; 


when 

s 

II 
1— » 

03 

^ changes  sign  from  — to  + 

a minimum  ; 

when 

2 = 3, 

^ changes  sign  from  + to  — 

a maximum ; 

hence  when 

x = 13, 

u — -tf-5-,  a minimum  value  ; 

and  when 

2 = 3, 

u = oo , a maximum  value. 

262 


EXAMPLES. 


7.  Examine  u = b + (x  — a)*  for  maxima  and  minima 
— = | (x  — a)%  = 0 ; x — a and  u — b 


When  x = a,  ^ changes  sign  from  — to  +. 


x = a gives  u = b,  a minimum. 

8.  Examine  u — b — (a  — x)*5  for  maxima  and  minima. 
du 


dx 


= f(«  — xY»  = °; 

du 


x — a and  u = b. 


When  x = a,  changes  sign  from  + to  — . 


x = a gives  u = b,  a maximum. 

9.  Examine  u = b + V a3  — 2a2x  + ax 2 for  maxima  and 
minima. 

If  u is  a maxima  or  minima,  u — b •will  be  so;  therefore 
we  omit  the  constant  b and  the  radical  by  Art.  99,  and  get 

u'  — a3  — 2a2x  + ax 2 ; 

~ — 2a2  + 2 ax  = 0;  .'.  x — a and  u = b. 

ax 

du' 

When  x = a,  changes  sign  from  — to  4 . 
x = a gives  u — b>  a minimum. 

cfix 

10.  Examine  u = ^ ^ for  maxima  and  minima 

Using  the  reciprocal,  since  it  is  more  simple,  and  omitting 
the  constant  a 2 (Art.  99),  we  have 

(a  — x)2  a2 


EXAMPLES. 


163 


x — ± a,  and 


cPu'  _ 2 

dx 2 a 


Hence,  x — + a makes  u'  a minimum,  and  x = — a 
makes  it  a maximum ; therefore,  since  maxima  and  minima 
values  of  u'  correspond  respectively  to  the  minima  and 
maxima  values  of  u (Art.  99,  4),  we  have, 


when  x = a,  u = <x> , a maximum. 


x = — a,  u 


a . . 

- , a minimum. 


Find  the  values  of  x which  give  maximum  and 
minimum  values  of  the  following  functions : 


1. 


2. 


3. 

4. 

6. 

6. 


7. 


8. 

9. 

10. 

11. 


_ u = Xs  — 3a;2  — 24a;  + 85. 

Ans.  x = — 2,  max. ; x = 4,  min. 
u = — 21a;2  + 36a;  — 20. 

x = 1,  max.;  x = 6,  min. 


u = x3  — 18a:2  + 96a;  — 20. 

x = 4,  max. ; 

(a  — x)s 

u — ± -A. 

a — 2x 


x = 8,  min. 
x = {a,  min 


u 


1 + 3x 


x = — l -fc,  max. 


a/4  + 5a; 
u — x3  — 3a;2  — 9a;  + 5. 

x — —l,  max. ; x = 3,  min. 
u = Xs  — 3a?  + 6a;  + 7. 

Neither  a max.  nor  a mm. 
u = (x  — 9)5  (x  — 8)4. 

x — 8,  max. ; x = 8$- , min 
x 


u = 


x = cos  a;,  max. 


1 + x tan  x 
u = sin3  a;  cos  a;,  x = 60°,  max. 
sin  x 


u — 


1 -h  tan  x 


x - 45°,  max. 


164 


GEOMETRIC  PROBLEMS. 


12. 

13. 


u = sin  x + cos  x. 

x — 45°,  max. ; 


u 


log  X 
xn 


x = 225°,  mm. 

1 

x — en,  max. 


GEOMETRIC  PROBLEMS. 

The  only  difficulty  in  the  solution  of  problems  in  maxima  ana 
minima  consists  in  obtaining  a convenient  algebraic  expression  for  the 
function  whose  maximum  or  minimum  value  is  required.  No  gen- 
eral  rule  can  well  be  given  by  which  this  expression  can  be  found. 
Mucli  will  depend  upon  the  ingenuity  of  the  student.  A careful  ex- 
amination of  all  the  conditions  of  the  problem,  and  tact  in  applying 
his  knowledge  of  principles  previously  learned  in  Algebra,  Geometry, 
and  Trigonometry,  with  experience,  will  serve  to  guide  him  in  form- 
ing the  expression  for  the  function.  After  reducing  the  expression  to 
its  simplest  form  by  the  axioms  of  Art.  99,  he  must  proceed  as  in 
Art.  97. 

1.  Find  the  maximum  cylinder  which 
can  be  inscribed  in  a given  right  cone 
with  a circular  base. 

Suppose  a cylinder  inscribed  as  in 
the  figure.  Let  AO  = i,  DO  = a, 

CO  = x,  CE  = y. 

Then,  denoting  the  volume  of  the 
cylinder  by  v,  we  have 

V = 7 ry2x. 

From  the  similar  triangles  DOA  and  DCE,  we  have 
DO  : AO  ::  DC  : EC, 
m a : l : : a — x : y; 

***  y ~ \ “ *)» 
v — (a  — z)*x. 


(1) 


which  in  (1)  gives 


(2] 


GEOMETRIC  PROBLEMS. 


165 


Dropping  constant  factors  (Art.  99),  we  have 


or 


u = (a  — a;)2  x = a2x  — 2 ax2  4-  Xs ; 

^ = a2  — 4 ax  4-  3a;2  — 0, 
dx 


& — \ax  = — \a 2 ; 

S = -4a+te 


x — a 


= 2a,  when  x — a, 
= — 2a,  when  a;  = \a, 


or  $a. 

minimum; 

maximum. 


Hence  the  altitude  of  the  maximum  cylinder  is  one-third 
of  the  cone. 

The  second  value  of  x in  (2)  gives 

TfS3,  , sd  , TO 

V ~ (® 


Volume  of  cone  = \nab2. 

Volume  of  cylinder  r=  $ volume  of  cone. 

y = - (a  — |a)  =.  \b  — radius  of  base  of  cylinder. 
$ 


2.  What  is  the  altitude  of  the 
maximum  rectangle  that  can  be  in- 
scribed in  a given  parabola  ? 

Let  AX  = a,  AH  = a-,  DH  = y, 
and  A = area  of  rectangle.  Then 
we  have 


A 


D 

/ H 

\E 

/ 

\ 

B M X N Q 

Fig.  II. 


A — 2y  (a  — x). 


(1) 


But  from  the  equation  of  the  parabola,  we  have 
y = V2 px, 

which  in  (1)  gives  A = 2 V2px  (a  — x).  (2) 

u'  = V x (a  — x)  = ax ? — x%. 

— \az~*  — fa;?  = 0.  x = \a. 


166 


GEOMETRIC  PROBLEMS. 


cl 'LL 

Since  this  value  of  x makes  change  sign  from  + to 

— , it  makes  the  function  A a maximum;  therefore  the 
altitude  of  the  maximum  rectangle  is  § a. 

3.  What  is  the  maximum  cone  that 
can  be  inscribed  in  a given  sphere  ? 

Let  ACB  be  the  semicircle,  and 

7 A 

ACD  the  triangle  which,  revolved  Fi  l2_  0 

about  AB,  generate  the  sphere  and 
cone  respectively.  Let  AO  = r,  AD  = x,  and  CD  = y, 
and  v — volume  of  cone. 


Then  v — \~y2x. 

But  y 2 = AD  x DB  = (2 r — x)  x, 

which  in  (1)  gives  v = (2 r — x)  x 2, 

or  u = 2rx 2 — x3 ; 

du 


(1) 

(2) 


dx 


= 4.rx  — 3 a;2  - - 0. 


x 


0 and  |r. 


du 


n 


The  latter  makes  — change  sign  from  + to  — ; 
makes  v a maximum. 

Hence  the  altitude  of  the  maximum  cone  is  f of  the 
diameter  of  the  sphere. 

The  second  value  of  x in  (2)  gives 

v = i77  (2r  — $r)  (|r)2  — ff"^  = x i-r*. 

V olume  of  sphere  = |-Trs ; 

the  cone  = ^ of  the  sphere. 

4.  Find  the  maximum  parabola  which  can  he  cut  from  a 
given  right  cone  with  a circular  base,  knowing  that  the  area 
of  a parabola  is  f the  product  of  its  base  and  altitude. 


GEOMETRIC  PROBLEMS. 


167 


Let  AB  rr:  a,  AC  = b,  and  BH  = x\ 
then  AH  = a — x. 

FE  = 2EH  = 2a/AH/TBH 
— 2a/  (a  — x)x. 

Also,  BA  : AC  ::  BH  : HD, 

or  u : b ::  x : HD  = - x. 

a 


Calling  the  parabola  A,  we  have 

A — |FE  x HD  = 4 -a;\/ (a  — x)  x, 
a ' 


or  u = ax3  — x4. 

~ - 3 ax2  — 4x3  = 0; 

x — 0 and  x = \a. 

The  second  value  makes  ^ change  sign  from  + to  — , 
and  makes  the  function  A a maximum. 

A — |«a/ (a  — fa)  fa  = \abV3, 

which  is  the  area  of  the  maximum  parabola. 

Rem. — In  problems  of  maxima  and  minima,  it  is  often  more  con- 
venient to  express  the  function  u in  terms  of  two  variables,  x and  y, 
which  are  connected  by  some  equation,  so  that  either  may  be  regarded 
as  a function  of  the  other.  In  this  case,  either  variable  of  course  may 
be  eliminated,  and  u expressed  in  terms  of  the  other,  and  treated  by 
the  usual  process,  as  in  Examples  1,  2,  and  3.  It  is  often  simpler, 
however,  to  differentiate  the  function  u,  and  the  equation  between  x 
and  y,  with  respect  to  either  of  the  variables,  x,  regarding  the  other, 

y,  as  a function  of  it,  and  then  eliminate  the  first  derivative,  The 

dx 

second  method  of  the  following  example  will  illustrate  the  process. 


168 


GEOMETRIC  PROBLEMS. 


5.  To  find  the  maximum  rectangle  inscribed  in  a given 
ellipse. 


Let  CM  = x,  PM  = y,  and 
A — area  of  rectangle.  Then  we 
have 

A — 4 xy,  (1) 

and  a2y 2 + b2xz  — aW.  (2) 

1st  Method. — From  (2)  we  get 

y = - V a2  — 

a a 


Fig.  14. 


which  in  (1)  gives  A =4  ^x  y/ az  — x1, 

or  u — a2x2  — x4. 

^ = 2 a2x  — 4 x3  = 0.  .'.  x = ± -—7=-' 
ax  y2 

x = 4-  -^-=  makes  ^ change  sign  from  4-  to  — ; .-.it 
V2  dx 
makes  A a maximum. 

Hence,  the  sides  of  the  maximum  rectangle  are  a v 2 
and  J \/2,  and  the  area  is  2 ab. 


2d.  Method. — Differentiate  (1)  and  (2)  with  respect  to  x 
after  dropping  the  factor  4 from  (1),  and  get 


(LA 

dx 


dy  A 

= ’J  + X\ 5 = 0; 


2a2y 


dy 

dx 


-f  2 tLx  — 0; 


. dy=_y. 

dx  x 

dy  _ &x_ 

dx  — a2# 


^ ^ , or  lAx2  =■  a2;/2 ; 

a2^  x 


which  in  (2)  gives 


GEOMETRIC  PROBLEMS. 


169 


2a2i/  = a?b2, 


y 


— = and 

V* 


x = 


a/2 


6.  Find  the  cylinder  of  greatest  convex  surface  that 

can  be  inscribed  in  a right  circular  cone,  whose  altitude 

is  h and  the  radius  of  whose  base  is  r.  0 , t. hr 

surface  - — 

/y 


7.  Determine  the  altitude  of  the  maximum  cylinder 
which  can  be  inscribed  in  a sphere  whose  radius  is  r. 

Altitude  = | r a/3. 

8.  Find  the  maximum  isosceles  triangle  that  can  be 

inscribed  in  a circle.  An  equilateral  triangle. 

9.  Find  the  area  of  the  greatest  rectangle  that  can  be 
inscribed  in  a circle  whose  radius  is  r. 

The  sides  are  each  r a/2. 

10.  Find  the  axis  of  the  cone  of  maximum  convex  sur- 
face that  can  be  inscribed  in  a sphere  of  radius  r. 

The  axis  = | r. 

11.  Find  the  altitude  of  the  maximum  cone  that  can  be 

inscribed  in  a paraboloid  of  revolution,  whose  axis  is  a,  the 
vertex  of  the  cone  being  at  the  middle  point  of  the  base  of 
the  paraboloid.  Altitude  = 

12.  Find  the  altitude  of  the  cylinder  of  greatest  convex 
surface  that  can  be  inscribed  in  a sphere  of  radius  r. 

Altitude  = r a/2. 

13.  From  a given  surface  s,  a vessel  with  circular  base 
and  open  top  is  to  be  made,  so  as  to  contain  the  greatest 
amount.  Find  its  dimensions.  (See  Remark  under  Ex.  4.) 

The  altitude  = radiits  of  base  = y —• 

on 


14.  Find  the  maximum  cone  whose  convex  surface  is 
constant.  The  altitude  = a/2  times  the  radius  of  base. 

15.  Find  the  maximum  cylinder  that  can  be  inscribed  in 
an  oblate  spheroid  whose  semi-axes  are  a and  l. 

_ 2 

The  radius  of  base  — a the  altitude  = b — =• 


8 


170 


GEOMETRIC  PROBLEMS. 


16.  Find  the  maximum  difference  between  the  sine  and 

cosine  of  any  angle.  When  the  angle  = 135°. 

17.  Find  the  number  of  equal  parts  into  which  a must 
be  divided  so  that  their  continued  product  may  be  a 
maximum. 

Let  x be  the  number  of  parts,  and  thus  each  part  equals 
~,  and  therefore  u = , from  which  we  get  x — a ; 


therefore  each  part  = e,  and  the  product  of  all  = (e'y. 

18.  Find  a number  x such  that  the  xffi  root  shall  be  a 

maximum.  x — e — 2.71828 + . 

19.  Find  the  fraction  that  exceeds  its  mth  power  by  the 

greatest  possible  quantity.  / 1 \-tr 

\ml 


20.  A person  being  in  a boat  3 miles  from  the  nearest 
point  of  the  beach,  wishes  to  reach  in  the  shortest  time  a 
place  5 miles  from  that  point  along  the  shore  ; supposing 
he  can  walk  5 miles  an  hour,  but  row  only  at  the  rate  of 
4 miles  an  hour,  required  the  place  he  must  land. 

One  mile  from  the  place  to  be  reached. 

21.  A privateer  wishes  to  get  to  sea  unmolested,  but  has 
to  pass  between  two  lights,  A and  B,  on  opposite  head- 
lands, the  distance  between  which  is  c.  The  intensity  of 
the  light  A at  a unit’s  distance  is  a,  and  the  intensity  of  B 
at  the  same  distance  is  i ; at  what  point  between  the  lights 
must  the  privateer  pass  so  as  to  be  as  little  in  the  light  as 
possible,  assuming  the  principle  of  optics  that  the  intensity 
of  a light  at  any  distance  equals  its  intensity  at  the  distance 
one  divided  by  the  square  of  the  distance  from  the  light. 

ca? 
a?  + 

22.  The  flame  of  a candle  is  directly  over  the  centre  of  a 
circle  whose  radius  is  r ; what  ought  to  be  its  height  above 
the  plane  of  the  circle  so  as  to  illuminate  the  circumfer- 
ence as  much  as  possible,  supposing  the  intensity  of  the 


GEOMETRIC  PROBLEMS. 


171 


light  to  vary  directly  as  the  sine  of  the  angle  under  which 
it  strikes  the  illuminated  surface,  and  inversely  as  the 
square  of  its  distance  from  the  same  surface. 

Height  above  the  plane  of  the  circle  = r Vi- 

23.  Find  in  the  line  joining  the  centres  of  two  spheres, 
the  point  from  which  the  greatest 
portion  of  spherical  surface  is 
visible. 

The  function  to  be  a maximum 
is  the  sum  of  the  two  zones  whose 
altitudes  are  AD  and  ad ; hence 
we  must  find  an  expression  for  the  areas  of  these  zones. 

Put  CM  = R and  cm  = r,  C c — a and  CP  = x. 

The  area  of  the  zone  on  the  sphere  which  has  R for  its 
radius  (from  Geometry,  or  Art.  194)  = 2-RAD  ==  2ttR2 
/ R3\ 

— 2uRCD  = 2-  ( R2  — — ),  and  in  the  same  way  for  the 
other  zone,  from  which  we  readily  obtain  the  solution. 


Fig, 15. 


X — 


aR« 


Rt  + r* 

24.  Find  the  altitude  of  the  cylinder  inscribed  in  a sphere 
of  radius  r,  so  that  its  whole  surface  shall  be  a maximum. 

Altitude  = r 2 (l  — 


CHAPTER  IX. 


TANGENTS,  NORMALS  AND  ASYMPTOTES. 


J.00.  Equations  of  the  Tangent  and  Normal — Let 

P,  (x\  y)  be  the  point  of  tangency  ; 
the  equation  of  the  tangent  line  at 
(; x ',  y’)  will  be  of  the  form  (Anal. 

Geom.,  Art.  25) 

y — y = a (x  - x’),  (i) 

in  which  a is  the  tangent  of  the 
angle  which  the  tangent  line  makes 
with  the  axis  of  x.  It  was  shown  in 
Article  56a  that  the  value  of  this  tangent  is  equal  to  the 
derivative  of  the  ordinate  of  the  point  of  tangency,  with 
J espect  to  x, 


or 


Hen  je 


y-y' 


dx1 


(x  — x’), 


(2) 


is  the  equation  of  the  tangent  to  the  curve  at  the  point 
(: xr , y'),  x and  y being  the  current  co-ordinates  of  the 
tangent. 

Since  the  normal  is  perpendicular  to  the  tangent  at  the 
point  of  tangency,  its  equation  is,  from  (2), 


y-y' 


dx' 

dy' 


[x  - x). 


(Anal.  Geom.,  Art.  27,  Cor.  2.) 


(3) 


EXAMPLES. 


173 


Rem. — To  apply  (2)  or  (3)  to  any  particular  curve,  we 
dy1  doc' 

substitute  for  ~ or  j—, , its  value  obtained  from  the  equa- 

CIX  CL  U 

tion  of  the  curve  and  expressed  in  terms  of  the  co-ordinates 
of  the  point  of  tangency. 


EXAMPLES. 


1.  Find  the  equations  of  the  tangent  and  normal  to  the 
ellipse 

a2y2  + b2x2  — a2IP. 


dy 

dx 


We  find 
and  this  value  in  (2)  gives, 


h2x 


a2y’ 


df 

dx' 


h2x' 

«y; 


y-y  = 

which  by  reduction  becomes, 

a2yy'  -f  l2xx’  = cfiV2, 
which  is  the  equation  of  the  tangent ; and 


, dhg 

y-y 


is  the  equation  of  the  normal. 


2.  Find  the  equations  of  the  tangent  and  normal  to  the 
parabola  if  = 2px. 

dy  _ p dy'  _ p 

dx  ~~  y’  dx'  ~ y” 


We  find 
and  this  value  in  (2)  gives 


or 

But 


y-y'  = f (x  - x')> 

yy'  — y'*  = px—px'. 


y'  2 = 2 px'  ; 


174 


EXAMPLES. 


•••  yy'  — p(x  + x'), 
which  is  the  equation  of  the  tangent j and 

y - y'  = ~ | (x  ~ *') 

is  the  equation  of  the  normal. 

3.  Find  the  equations  of  the  tangent  and  normal  to  an 
hyperbola. 

Tangent,  a2yy'  — Vxx  =.  — a2h2. 

Normal,  y — y'  = — ^{x  — x'). 


4.  Find  the  equation  of  the  tangent  to  3 y2  + x2  — 5 = 0, 
at  x = 1. 

Here  = } -=  = =F  -29  about, 

dx  by  ± 3.465 

which  in  (2)  gives 

yT  1.155  = q=.29(a  — 1), 

or  y — T -29.r  ± 1-44 

Hence  there  are  two  tangents  to  this  locus  at  x = 1, 
their  equations  being 

y = — .29a  + 1.44  and  y = 4-  .29a:  — 1.44. 


5.  Find  the  equation  of  the  tangent  to  the  parabola 
y 2 = 9a,  at  x — 4. 

At  (4,  6)  the  equation  is  y = \x  + 3. 

“ (4,  — 6)  “ « “ y = — \x  — 3. 

6.  Find  the  equation  of  the  normal  to  y 2 = 2a-2  — a-3,  at 

x = 1. 

At  (1,  + 1)  the  equation  is  y = — 2a  + 3. 

“ (1}  _ i)  “ “ “ y=  2a  — 3. 

7.  Find  the  equation  of  the  normal  to  y2  — Gx  — 5,  at 

y — 5,  and  the  angle  which  this  normal  makes  with  the 
axis  of  x.  y = — $a  + ; angle  = tan"1  (—  f). 


LENGTH  OF  TANGENT,  NORMAL,  ETC. 


175 


101.  Length  of  Tangent,  Normal.  Subtangent, 
Subnormal,  and  Perpendicular  on  the  Tangent  from 
the  Origin. 

Let  PT  represent  the  tangent  at 
the  point  P,  PiST  the  normal;  draw 
the  ordinate  PM  ; then 

Y 

1 

MT 

MN 

is  called  the  subtangent, 

“ “ “ subnormal.  ° 

P 

J 

Let 

a — angle  PTM;  / 

V M N " 

° Fig.  17. 

then  tan  a = (Art.  56a). 

1st. 

TM  = MP  cot  a 

, dec* 

Subtangent  — y' 

II 

a,i  a. 

2d. 

MN  = MP  tan  MPN  = y'  tan  «; 

Subnormal  — 

3<L 

PT  = VPM2  + MT2 

ii 

< 

to 

+ 

, dx\> 
ydy)> 

Tangent  = y' 1 + 1 

\dy) 

ith» 

PN  = VPM2  + MN2 

= \A'!  + ( 

V'dy'\\ 
J dx ) ’ 

•••  Normttl  = ^V1  + (*?)' ' 


5th.  The  equation  of  the  tangent  at  P (x',  y')  is  (Art.  100), 


176 


EXAMPLES. 


y-y  = s{x-x^ 

or  xdy'  — ydx  — x'dy'  + y'dx'  = 0 ; 

which,  written  in  the  normal  form,  is 

xdy'  — ydx'  — x'dy'  -f  y'dx 


hence,  OD 


V\dx ')2  + ( dy')2 

y'dx  — x'dy' 


(Anal.  Geom.,  Art.  24) 


V ( dx')2  -\-  ( dy 'f 
Perpendicular  on  the  tangent  f rom  the  origin 
^ y'dx'  — x'dy' 


V ( dx')2  + (dy')2 

Sch. — In  these  expressions  for  the  subtangent  and  sub- 
normal it  is  to  be  observed  that  the  subtangent  is  measured 
from  M towards  the  left,  and  the  subnormal  from  M towards 

dy 

the  right.  If,  in  any  curve,  y'  fx  *s  a negative  quantity,  it 

denotes  that  N lies  to  the  left  of  M,  and  as  in  that  case 
fix' 

y'  ^-7  is  also  negative,  T lies  to  the  right  of  M. 


EXAMPLES. 


1.  Find  the  values  of  the  subtangent,  subnormal,  and 
perpendicular  from  the  origin  on  the  tangent,  in  the  ellipse 
<py2  b2x2  — a2b2. 


Here 


dy'  _ h2x' 

dx'  ~ a2y' 


Hence, 


the  subtangent  = y 


, dx' 
dy’ 


a2y'2 

W’ 


the  subnormal 


df 

dx' 


EXAMPLES. 


m 


the  perpendicular  from  origin  on  tangent 

_ aW 
~ («Y 2 + 6V2)T 

2.  Find  the  subtangent  and  subnormal  to  the  Cissoid 
x3 


y 2 = 


2a  — x 


Here 


(See  Anal.  Geom.,  Art.  149.) 

dy'  _ xi  (3 a — x) 
dx  (2a  — x)  * 


Hence,  the  subtangent  = 


x (2a  — x) 
3 a — x 


the  subnormal  = xJ^a ^ . 

(2a  — x)2 

3.  Find  the  value  of  the  subtangent  of  y 2 = 3a:2  — 12. 

at  x — 4.  Subtangent  = 3. 

4.  Find  the  length  of  the  tangent  to  y2  = 2x,  at  x — 8. 

Tangent  = 4\/l7. 

5.  Find  the  values  of  the  normal  and  subnormal  to  the 
cycloid  (Anal.  Geom.,  Art.  156). 


x — r vers-1  - — a/ 2 ry 


y2‘, 


dx 


\/2ry—y2 


" dy  V 2ry-y 2 %r  — y 

dy  _ 2 r — y 

dx  V 2 ry  — y2 


Fig.  18. 


Subnormal  = V2  ry  — y2  — MO. 

Normal  = \/2ry  = PO. 

It  can  be  easily  seen  that  PO  is  normal  to  the  cycloid  at 
P ; for  the  motion  of  each  point  on  the  generating  circle  at 


178 


POLAR  CURVES. 


the  instant  is  one  of  rotation  about  the  point  of  contact  0, 
he.,  each  point  for  an  instant  is  describing  an  infinitely 
small  circular  arc  whose  centre  is  at  0 ; and  hence  PO  is 
normal  to  the  curve,  he.,  the  normal  passes  through  the 
foot  of  the  vertical  diameter  of  the  generating  circle.  Also, 
since  OPII  is  a right  angle,  the  tangent  at  P passes  through 
the  upper  extremity  of  the  vertical  diameter. 

6.  Find  the  length  of  the  normal  in  the  cycloid,  the 
radius  of  whose  generatrix  is  2,  at  y — 1.  Normal  = 2. 

POLAR  CURVES. 

102.  Tangents,  Normals,  Subtangents,  Subnor- 
mals, and  Perpendicular  on  Tangents. 

Let  P be  any  point  of  the 
curve  APQ,  0 the  pole,  OX  the 
initial  line.  Denote  XOP  by 
d,  and  the  radius-vector,  OP, 
by  r.  Give  XOP  the  infinitesi- 
mal increment  POQ  = dd,  then 
OQ  = r + dr.  From  the  pole 
0,  with  the  radius  OP  = r,  de- 
scribe the  small  arc  PR,  sub- 
tending dd ; then,  since  dd  = ab 
is  the  arc  at  the  unit’s  distance 
from  the  pole  0,  we  have 

PR  = rdd  and  RQ  = dr.  (1) 

Let  PQ,  the  element*  of  the  arc  of  the  curve,  be  repre- 
sented by  ds. 

.*.  PQ2  = Hi2  + RQ2, 

or  ds'  — r2  d(f  + di2.  (2) 

Pass  through  the  two  points  P and  Q the  right  line  QPT; 


* See  Art.  56 a.  foot-note. 


POLAR  CURVES. 


179 


then,  as  P and  Q are  consecutive  points,  the  line  QP  T is  a 
tangent  to  the  curve  at  P (Art.  56 a).  Through  P draw  the 
normal  PC,  and  through  0 draw  COT  perpendicular  to  OP, 
and  OD  perpendicular  to  PT.  The  lengths  PT  and  PC  are 
respectively  called  the  polar  tangent  and  the  polar  normal. 
OC  is  called  the  polar  subnormal ; OT  the  polar  subtangent ; 
aud  OD,  the  perpendicular  from  the  pole  on  the  tangent,  is 
usually  symbolized  by  p.  The  value  of  each  of  these  lines  is 
required. 

tanEQP  = ?t  = 5,from(l).  (3) 


Since  OPT  = OQT  -j-  dd,  the  two  angles  OPT  aud  OQT 
differ  from  each  other  by  an  infinitesimal,  and  therefore 
OPT  = OQT,  and  hence, 


Vd6 

tan  OPT  = from  (3), 


(4) 


~D  T)  rdf) 

sin  OPT  sin  OQP  = 7^  - — , from  (1).  (5) 

Qr  as 


Hence, 


OT  = 

= polar 

subtangent 

= OP 

’ tan  OPT 

II 

1 

from  (4). 

(6) 

OC  = 

= polar 

subnormal 

= OP 

tan  OPC 

= OP  cot  OPT 

II 

tll^ 

, from  (4). 

(7) 

PT  = 

= polar 

tangent  = 

VoP‘ 

+ OT2  = 

r f / 1 + r2 

eld* 
dr 2’ 

from  (6). 

(8) 

PC  = 

= polar 

normal  — 

Vop2 

+ 

Oi 

O 

to 

II 

J 2 , dr 2 

from  (7). 

(9) 

180 


EXAMPLES. 


OD  ■=  n — OP  sin  OPD  = — = — from  (5)  = . 

as  -y/ r2dd 2 + dr2'- 

from  (2).  (10) 

See  Price’s  Calculus,  Vol.  I,  p.  417. 

EXAMPLES. 

1.  The  spiral  of  Archimedes,  whose  equation  is  r = ad. 
(Anal.  Geom.,  Art.  160.) 

dd  1 7*2 

Here  — = - ■ Subt.  = — , from  (6). 

dr  a a 

Subn.  = a,  from  (7). 

Tangent  = r */r  + from  (8)> 

Normal  = \/r2  + a 3,  from  (9). 

P = » from  (10)- 

■yr2  + a2 

2.  The  logarithmic  spiral  r — ae.  (Anal.  Geom., 
Art.  163.) 


RECTILINEAR  ASYMPTOTES. 


181 


Tan.  OPT  = 


td6  _ _J_  _ 
dr  ~ log  a ’ 


which  is  a constant ; and  therefore  the  curve  cuts  every 
radius-vector  at  the  same  angle,  and  hence  it  is  called  the 
Equiangular  Spiral. 


If  a — e,  the  Naperian  base,  we  have, 


tan  OPT  = = 1,  and  .\  OPT  = 45°, 

log  e 

and  OT  = OP  = r. 


3.  Find  the  subtangent,  subnormal,  and  perpendicular  in 
the  Lemniscate  of  Bernouilli,  r2  = a2  cos  26.  (Anal. 
Geom.,  Art.  154.) 

^3 

Subtangent  = ^ ; 

° a 2 sm  20 


Subnormal 


a2 

r 


sin  20 ; 


Y 3 ^3 

Perpendicular  = — — WM  : = — • 

V r4  + sin2  20 


4.  Find  the  subtangent  and  subnormal  in  the  hyper- 
bolic* spiral  rO  = a.  (Anal.  Geom.,  Art.  161.) 

7-2 

Subt.  = — ■ a ; Subn.  = -• 

a 


RECTILINEAR  ASYMPTOTES. 

103.  A Rectilinear  Asymptote  is  a line  which  is 
continually  approaching  a curve  and  becomes  tangent  to  it 
at  an  infinite  distance  from  the  origin,  and  yet  passes 
within  a finite  distance  of  the  origin. 

To  find  whether  a proposed  curve  has  an  asymptote,  we 
must  first  ascertain  if  it  has  infinite  branches,  since  if  it 


* This  curve  took  its  name  from  the  analogy  between  its  equation  and  that  of 
the  hyperbola  xy  = a ■ (See  Strong’s  Calculus,  p.  145 ; also  Young’s  Dif.  Calculus, 
120.) 


182 


RECTILINEAR  ASYMPTOTES. 


has  not,  there  can  be  no  asymptote.  If  it  has  an  infinite 
branch,  we  must  then  ascertain  if  the  intercept  on  eithei 
of  the  axes  is  finite.  The  equation  of  the  tangent 
(Art.  100)  being, 

, dy’  , 

y-y  = &(*-*)> 

if  we  make  successively  y — 0,  x = 0,  we  shall  find  for 
the  intercepts  on  the  axes  of  x and  y,  the  following  : 

dx 

x'  = x-yTy, 

(by  putting  x = x0  and  y = y0,  and  dropping  accents), 

Ay 


y°  = y 


dx  ’ 


Now,  if  for  x — co  both  x0  and  y0  are  finite,  they  wifi 
determine  two  points,  one  on  each  axis,  through  which  an 
asymptote  passes.  If  for  y = co  , x0  is  finite  and  y0  infi- 
nite, the  asymptote  is  parallel  to  the  axis  of  y.  If  for 
x — oo , x0  is  infinite  and  y0  finite,  the  asymptote  is  parallel 
to  the  axis  of  x.  If  both  x0  and  y0  are  infinite,  the  curve 
has  no  asymptotes  corresponding  to  x = oo.  If  both  x0 
and  y0  are  0,  the  asymptote  passes  through  the  origin,  and 

dy 

its  direction  is  obtained  by  evaluating  for  x — co . 

When  there  are  asymptotes  parallel  to  the  axis,  they  may 
usually  be  detected  by  inspection,  as  it  is  only  necessary  to 
ascertain  what  values  of  x will  make  y — oo , and  what 
values  of  y will  make  x = oo  . For  example,  in  the  equa- 
tion xy  = m,  x — 0 makes  y = oo,  and  y — 0 makes 
x — oo  ; hence  the  two  axes  are  asymptotes.  Also  in  the 
equation  xy  — ay  — bx  — 0,  which  may  be  put  in  eithei 
of  the  two  forms, 

bx  ay 


V = 


x — a 


or  x — 


y — $ o when  x — a,  and 


y — b 

- oo  when  y = b ; 


EXAMPLES. 


183 


hence  the  two  lines  x = a and  y — b are  asymptotes  to 
the  curve. 

In  the  logarithmic  curve  y — a1, 

y — 0 when  x = — go  , 

therefore  the  axis  of  x is  an  asymptote  to  the  branch  in  the 
second  angle. 

Also  in  the  Cissoid  %fl  = - — — , 

^ 2a  — x 

y — oo  when  x — 2a ; 
hence  x = 2 a is  an  asymptote. 


EXAMPLES 


1.  Examine  the  hyperbola 

a2y2  — Vhfl  = — cfllfl,  for  asymptotes. 

Here 


cly  _ b2x  _ 
dx  ~~  (fly  ’ 


rfi 

x0  = x — = — = 0 for  x = ± <x> . 
blx  x 


Vx2 

y°=y--¥y 


— = 0 for  v = ± oo. 

y 


Hence  the  hyperbola  has  two  asymptotes  passing  through 
the  origin. 


Also 


cly  _ l)2x 
dx  cfly 


for  x — go  . 


Hence  the  asymptotes  make  with  the  axis  of  x an  angle 
whose  tangent  is  ± ^ ; that  is,  they  are  the  produced 
diagonals  of  the  rectangle  of  the  axes. 

2.  Examine  the  parabola  >f  = 2px  for  asymptotes. 


184  ASYMPTOTES  DETERMINED  BY  EXPANSION. 


Here 

dy  p y2  , 

-f-  — - ; .*.  x0  = — = — oo  when  x or  w = ao , 

dxy  2p  3 

y0  = | = oo  when  y = co  or  a;  = oo  . 

Hence  the  parabola  has  no  asymptotes. 

The  ellipse  and  circle  have  no  real  asymptotes,  since 
neither  has  an  infinite  branch. 


3.  Examine  y3  = ax2  -f-  x3  for  asymptotes. 

When  x — ± oo , y = oo  ; the  curve  has  two 
infinite  branches,  one  in  the  first  and  one  in  the  third 
angle. 

dy  _ 2ax  + 3a:2. 
dx  ~ 3y2  ’ 

3 y3  ax 2 a 

~ x~  2 ax  + 3x2  ~ ~ 2ax  + 3a;2  ~ ~ 3* 
when  x = oo. 


2 ax2  + 3X3  3 (y3  — x3)  — 2 ax? 

y*  = y — w~  = ~~ — w — 

ax2  a . 

= ; when  x = co. 

3 ( ax 2 + x3p  d 

Hence  the  asymptote  cuts  the  axis  of  a;  at  a distance 

— and  that  of  y at  a distance  ^ from  the  origin,  and  as 
o o 

it  is  therefore  inclined  at  an  angle  of  45°  to  the  axis  of  x, 

its  equation  is 

■ a 

y = * + r 

(See  Gregory’s  Examples,  p.  153.) 


104.  Asymptotes  Determined  by  Expansion. — A 

very  convenient  method  of  examining  for  asymptotes  con- 
sists in  expanding  the  equation  into  a series  in  descending 


EXAMPLES. 


185 


powers  of  x,  by  the  binomial  theorem,  or  by  Maclaurin’s 
theorem,  or  by  division  or  some  other  method. 


EXAMPLES. 

^3  _i_  CL'JlP’ 

1.  Examine  y 2 = — — — for  asymptotes. 

Then 

y — ± x \/x  + a = ± x (1  + - + + etc.)  (1) 

J f x — a v x 2x?  ’ v/ 

When  x ■=  oo  (1)  becomes 

y - ± (*  + «)•  (2) 

We  see  that  as  x increases,  the  ordinate  of  (1)  increases, 
and  when  x becomes  infinitely  great,  the  difference  between 
the  ordinate  of  (1)  and  that  of  (2)  becomes  infinitesimal; 
that  is,  the  curve  (1)  is  approaching  the  line  (2)  and 
becomes  tangent  to  it  when  x — oo  ; therefore,  y = ±(x-\-a) 
are  the  equations  of  two  asymptotes  to  the  curve  (1)  at 
right  angles  to  each  other. 

Another  asymptote  parallel  to  the  axis  of  y is  given  by 
x — a. 


2.  Examine  z3  — xy1  + ay 2 = 0 for  asymptotes. 
Here 


y=± 


a? 


x — a 


/,  a 3 a?  5u3  , \ 

= ^l1  + 25  + 8^  + l6^4'etC-j 

Hence,  y = ± (x  + are  the  equations  of  the  two 
wymptotes. 

By  inspection,  we  find  that  x = a is  a third  asymptote. 


x% 1 

3.  Examine  y 2 — x2  ^ — for  asymptotes. 


186 


EXAMPLES. 


Here 


y = dt  x are  the  two  asymptotes. 


105.  Asymptotes  in  Polar  Co-ordinates.— When 

the  curve  is  referred  to  polar  co-ordinates,  there  will  be  an 
asymptote  whenever  the  subtangent  is  finite  for  r — ec. 
Its  'position  also  will  be  fixed,  since  it  will  be  parallel  to  the 
radius-vector.  Hence,  to  examine  for  asymptotes,  we  find 
what  finite  values  of  6 make  r — oo ; if  the  corresponding 
cIO 

polar  subtangent,  r 2 — , which  in  this  case  becomes  the 

perpendicular  on  the  tangent  from  the  pole,  is  finite  or  zero, 
there  will  be  an  asymptote  parallel  to  the  radius-vector.  If 
for  r — oo  the  subtangent  is  oo , there  is  no  corresponding 
asymptote. 


1.  Find  the  asymptotes  of  the  hyperbola  ary2  — fix2  — 
— a2b2  by  the  polar  method. 

The  polar  equation  is 


Therefore  the  asymptotes  are  inclined  to  the  initial  line 


EXAMPLES 


a2  sm2  0 - & cos2  6 = - (1) 

V2 

When  r = oo , (1)  becomes,  tan2  d = -2 ; 


dd 


aW 


From  (1)  we  get 


dr  ’ r3  ( a 2 + b2)  sin  d cos  6 ’ 

2 dd ab  ( b 2 cos2  d — a2  sin2  d)?  . . 

J~r  ~~  (a2  + ¥)  sin  d cos  d ’ 


and 


EXAMPLES. 


187 


which  is  equal  to  0 when  0 = tan-1  ^ ± - J ; hence  both 
asymptotes  pass  through  the  pole. 

2.  Find  the  asymptotes  to  the  hyperbolic  spiral  rd  — a. 
(See  Anal.  Geom.,  Art.  161.) 

Here  r = % , .-.  r = oo , when  0 = 0. 

U 

dd  a dd 

-T-  = „ , and  r2  ~r  — — a. 


There  is  an  asymptote  therefore  which  passes  at  a distance 
a from  the  pole  and  is  parallel  to  the  initial  line. 

3.  Find  the  asymptotes  to  the  lituus  rd?  = a.  (Anal. 
Geom.,  Art.  162.) 

Here  r = r = oo  , when  0 = 0. 

6? 

dd  2 a2  , 0dd  i . . _ 

= and  r2-r-  = — 2 ad*  = 0,  when  0 = 0. 

dr  r3  dr 

Therefore  the  initial  line  is  an  asymptote  to  the  lituus. 

4.  Find  the  asymptotes  of  the  Conchoid  of  Nicomedes, 
r = p sec  0 + m.  (Anal.  Geom.,  Art.  151.) 

Here  r — oo  when  0 = ^ ; and  r2~  — p when  0 = 

2 dr  1 2 

Therefore  the  asymptote  cuts  the  initial  line  at  right 
angles,  and  at  a distance  p from  the  pole. 


EXAMPLES. 


1.  Find  the  equation  of  the  tangent  to  3y2 — 2x 2 — 10  = 0, 

at  x — 4.  Ana.  y = ± .7127a:  ± .8909. 

^3 

2.  Find  the  equation  of  the  tangent  to  y2  = j , at 

2*3:  X 

. 

y = 2x  — 2 and  y = — 2a:  + 2. 


188 


EXAMPLES. 


3.  Find  the  equation  of  the  tangent  to  the  Kaperian 

logarithmic  curve.  Ans.  y = y'  (x  — x'  + 1). 

4.  At  what  point  on  y = x?  — Zx2  — 24a:  + 85  is  the 
tangent  parallel  to  the  axis  of  x ? 

dy' 

[Here  we  must  put  = 0.  See  Art.  56a.] 

At  (4,  5)  and  (—2,  113). 

5.  At  what  point  on  y2  = 2a4  does  the  tangent  make 

with  the  axis  of  x an  angle  whose  tangent  is  3,  and  where  is 
it  perpendicular?  At  (2,  4) ; at  infinity. 

6.  At  what  angle  does  the  line  y = \x  + 1 cut  the  curve 

y2  = 4a;  ? [Find  the  point  of  intersection  and  the  tangent 
to  the  curve  at  this  point;  then  find  the  angle  between  this 
tangent  and  the  given  line.]  10°  14'  and  33°  4'. 

7.  At  what  angle  does  y 2 = 10z  cut  x2  + y2  = 144  ? 

71°  0'  58". 

8.  Show  that  the  equation  of  a perpendicular  from  the 
focus  of  the  common  parabola  upon  the  tangent  is 

v' 

y = — -p  (*  - \p)- 

9.  Show  that  the  length  of  the  perpendicular  from  the 
focus  of  an  hyperbola  to  the  asymptote  is  equal  to  the  semi- 
conjugate axis. 

10.  Find  the  abscissa  of  the  point  on  the  curve 

y (x  — 1)  (a;  — 2)  = x — 3 

at  which  a tangent  is  parallel  to  the  axis  of  x. 

X = 3 

11.  Find  the  abscissa  of  the  point  on  the  curve 

y3  = (x  — a)2  (x  — c) 

at  which  a tangent  is  parallel  to  the  axis  of  x. 

2c  + a 


x = 


3 


EXAMPLES. 


189 


12.  Find  the  subtangent  of  the  curve  y = 


Ans. 


V2  a — x 
x (2a  — x ) 


3 a — x 

13.  Find  the  subtangent  of  the  curve  y3—3axy-\-x3  — 0. 

2 axy  — x3 
ay  — x1 

14.  Find  the  subtangent  of  the  curve  xy 2 — a2(a  — x). 

2 (ax  — x2) 
a 

15.  Find  the  subnormal  of  the  curve  y2  — 2 a2  log  x. 

a2 


16.  Find  the  subnormal  of  the  curve  3 ay2  + a3  = 2x?. 

x2 

a 

17.  Fin  a the  subtangent  of  the  curve  y2  = -• 

2x  (a  — x) 

3 a — 2x 

18.  Find  the  subtangent  of  the  curve 

x2y2  = (a  + x)2  (b2  — x2). 

x (a  + x)  (b2  — x2) 
x3  + ab2 

19.  Find  the  subnormal,  subtangent,  normal,  and  tangent 
n the  Catenary 


* = l(*7+*'7)- 


Subnormal  — -\ec  — e c ) : normal  = — • 


Subtangent 


c/s  -S\ 

= v; 

cy  . 


_ y 2 ' 


y2 

tangent  — -• 

(f  _ c2)v 


190 


EXAMPLES. 


20.  Find  the  perpendicular  from  the  pole  on  the  tangent 

in  the  lituus  r0%  = a.  2a2r 

V — r 

(r4  + 4 a4)^ 

21.  At  what  angle  does  y 2 = 2 ax  cut  a-3— 3 axy  + y3  = 0? 

cot-1  ^4. 

22.  Examine  y2  = 2x  + 3a:2  for  asymptotes. 

y = x/3  x + --=  is  an  asymptote. 

23.  Examine  y3  = 6x2  + x3  for  asymptotes. 

y — x + 2 is  an  asymptote. 

24.  Find  the  asymptotes  of  y2  (x  — 2 a)  — x?  — a3. 

x — 2a  ; y = ± (x  + a). 

25  Find  the  asymptotes  of  y — 

x = b ; x — 2b  ; y = x — 3 (a  — b) 


CHAPTER  X. 


DIRECTION  OF  CURVATURE,  SINGLE  POINTS, 
TRACING  OF  CURVES. 


106.  Concavity  and  Convexity. — The  terms  concav- 
ity and  convexity  are  used  in  mathematics  in  their  ordinary 
sense.  A curve  at  a point  is  concave  towards  the  axis  of  x 
when  in  passing  the  point  it  lies  between  the  tangent  and! 
the  axis.  See  Fig.  21.  It  is  convex  towards  the  axis  of  s 
when  its  tangent  lies  between  it  and  the  axis.  See 
Fig.  22. 

If  a curve  is  concave  down- 
wards, as  in  Fig.  21,  it  is  ]ilain 
that  as  x increases,  « decreases, 
and  hence  tan  a decreases ; that 

is,  as  x increases,  ~ (Art.  56«) 
decreases ; and  therefore  the  de- 
rivative of  (-~  or  (~-  is  negative. 
dx  dx2  ° 

In  the  same  way  if  the  curve  is 
convex  downward,  see  Fig.  22,  it  is 
plain  that  as  x increases,  a in- 
creases, and  therefore  tan  a in- 
creases ; that  is,  as  x increases, 

~ increases,  and  therefore  the  de- 
dx 


' 

f 

! 

/ /v 
Kc u/L\ 

T 

t m rvr 

(Art.  12.) 


dy 

dx 


dx 2 


is  positive. 


Hence  the  curve  is  concave  or  convex  downward  according 
dhj  . 

as  is  — or  -j-. 

dx2 


192 


POLAR  CO-ORDINATES. 


This  is  also  evident  from  Fig.  23,  where  MM'  =r  M'M'1 
= dx\  PP'  is  common  to  the  two  curves  and  the  common 
tangent.  PR  = P'R'  — dx ; and  P'R 
— P,R'.  But  P"R'  > P2R'  > P,R'. 

Now  P'R  and  P"R'  are  consecutive 
values  of  dy  in  the  upper  curve, 
and  P'R  and  P,R'  are  consecutive 
values  of  dy  in  the  lower  curve,  and 
hence  P"R'  — P'R  — d (dy)  = d2y  is 
-f , and  P,R'  — P'R  = d2y  is  — ; that 
is,  dzy  is  — or  +,  according  as  the 
curve  is  concave  or  convex  downwards. 
dtii 

The  sign  of  is  of  course  the  same  as  that  of  d2y, 
since  dx 2 is  always  positive. 

We  have  supposed  in  the  figures  that  the  curve  is  above 
the  axis  of  x.  If  it  be  below  the  axis  of  x,  the  rule  just 
given  still  holds,  as  the  student  mav  show  by  a course  of 
reasoning  similar  to  the  above. 

d%y 

If  the  curve  is  concave  downwards,  ■—  is  — ; if  it  be 


(fey 

above  the  axis  of  x,  y is  + ; therefore,  y ^ is  — ; if  the 


([2y 

curve  be  concave  upwards,  is  +;  if  it  be  below  the 


axis  of  x,  y is  — ; therefore,  y 


d2V  • 


d2y 


dx 2 


is  — ; that  is,  y 

3 dx1 


is  — when  the  curve  is  concave  towards  the  axis  of  x.  In 

(fiy 

the  same  way  it  may  be  shown  that  y is  -f , when  the 
curve  is  convex  towards  the  axis  of  x. 


107.  Polar  Co-ordinates. — A curve  referred  to  polar 
co-ordinates  is  said  to  be  concave  or  convex  to  the  pole  at 
any  point,  according  as  the  curve  in  the  neighborhood  of 
that  point  does  or  does  not  lie  on  the  same  side  of  the  tan- 
gent as  the  pole. 


EXAMPLES. 


193 


therefore  --  is  negative. 


If  therefore 


It  is  evident  from  Fig.  24,  that  when  the  curve  is  con- 
cave toward  the  pole  0,  as  r increases  p increases  also,  and 
civ 

therefore  ^ is  positive ; and  if  the  curve  is  convex  toward 

the  pole,  as  r increases  y?  decreases,  and 
dr 
dp 

the  equation  of  the  curve  is  given  in 
terms  of  r and  6,  to  find  whether  the 
curve  is  concave  or  convex  towards  the 
pole,  we  must  transform  the  equation 
into  its  equivalent  between  r and  p,  by 

means  of  (10)  in  Art.  102,  and  then  find  ^ 


Fig-  24i 


EXAMPLES. 

1.  Find  the  direction  of  curvature  of 
{x  — 1)  (a;  — 3) 


Here 


y = 

cPy 


x — 2 
2 


[x  - 2 )3  ’ 


(py  ■ 


that  is,  is  positive  or  negative,  according  as  x < or  >2-, 

and  therefore  the  curve  is  convex  downward  for  all  values 
of  x < 2,  and  concave  downwards  for  all  values  of  x > 2„ 


2.  Find  the  direction  of  curvature  of 

y — b + c (x  + a)2  and  y = a2  Vx  — a. 

A ns.  The  first  is  concave  upward,  the  second  is  concave 
towards  the  axis  of  x. 

3.  Find  the  direction  of  curvature  of  the  lituus  r = —• 

d v 


194 


SINGULAR  POINTS. 


dr  r r3  _ dr2 r6 

Here  dd^~  2d=~2d2’  W~  4^’ 

which  in  (10)  of  Art.  102  gives, 

_ 2 a2r  _ t rfr (4a4  + r4)^ 

^ ~ (r4  + 4a4)4’  " dp  ~ 2a2 (4a4  — r4)' 

Therefore  the  curve  is  concave  toward  the  pole  for  values 
of  r < a \/2,  and  convex  for  r > a *f2. 

4.  Find  the  direction  of  curvature  of  the  logarithmio 
spiral  r = a9. 

By  Art.  102,  Ex.  2, 

mr  _ _ r/r a/wi2  + 1 _ 

P ~ Vm2  + 1 dp  ~ m 

which  is  always  positive,  and  therefore  the  curve  is  always 
concave  toward  the  pole. 

SINGULAR  POINTS. 

108.  Singular  Points  of  a curve  are  those  points 
which  have  some  property  peculiar  to  the  curve  itself,  and 
not  depending  on  the  position  of  the  co-ordinate  axes. 
Such  points  are  : 1st,  Points  of  maxima  and  minima  ordi- 
nates ; 2d,  Points  of  inflexion  ; 3d,  Multiple  Points  ; 4th, 
Cusps ; 5th,  Conjugate  points ; 6th,  Stop  points ; 7th, 
Shooting  points.  We  shall  not  consider  any  examples  of 
the  first  kind  of  points,  as  they  have  already  been  illus- 
trated in  Chapter  VIII,  but  wdll  examine  very  briefly  the 
others. 

109.  Points  of  Inflexion. — A point  of  inflexion  is  a 
point  at  which  the  curve  is  changing  from  convexity  to 
concavity,  or  the  reverse  ; or  it  may  be  defined  as  the 
point  at  which  the  curve  cuts  the  tangent  at  that  point. 

When  the  curve  is  convex  downwards,  is  + (Art. 

dx~ 


EXAMPLES. 


195 


(fly 

106),  and  when  concave  downwards,  is  — ; therefore, 
cPy 

at  a point  of  inflexion  is  changing  from  + to  — , or 

from  — to  +,  and  hence  it  must  be  0 or  x.  Hence  to 

(fly 

find  a point  of  inflexion,  we  must  equate  to  0 or  oo , 

and  find  the  values  of  x ; then  substitute  for  x a value  a 

little  greater,  and  one  a little  less  than  the  critical  value  ; 

cfiy  . . . . . 

if  changes  sign,  this  is  a point  of  inflexion. 

(J.J0 


EXAMPLES. 

1.  Examine  y — b + {x  — a )3  for  points  of  inflexion. 

Here  § = 6 (*  — a)  = 0 ; 

x = a and  hence  y — b. 

This  is  a critical  point,  i.  e.,  one  to  be  examined  ; for  if 

there  is  a point  of  inflexion  it  is  at  x — a.  For  x > a, 

(fly  m cfly 

is  +,  and  for  x < a,  is  — . Hence  there  is  a point 

of  inflexion  at  (a,  b). 

2.  Examine  the  witch  of  Agnesi, 

x*y  = 4a2  (2a  — y), 
for  points  of  inflexion. 

There  are  points  of  inflexion  at  (±  ^r)- 

3.  Examine  y = b + [x  — a)5  for  points  of  inflexion. 

There  is  a point  of  inflexion  at  (a,  b). 

4.  Examine  the  lituus  for  points  of  inflexion. 

dv 

By  Art.  107,  Ex.  3,  is  changing  sign  from  4-  to  — 
when  r — a V"2,  indicating  that  the  lituus  changes  at  this 


196 


METHOD  OF  FINDING  MULTIPLE  POINTS. 


point  from  concavity  to  convexity,  and  hence  there  is  a 
point  of  inflexion  at  r = a \/2. 

110.  Multiple  Points. — A multiple  point  is  a point 
through  which  two  or  more  branches  of  a curve  pass.  If 
two  branches  meet  at  the  same  point,  it  is  called  a double 
point ; if  three,  a triple  point ; and  so  on.  There  are  two 
kinds  : 1st,  a point  where  two  or  more  branches  intersect, 
their  several  tangents  at  that  point  being  inclined  to  each 
other ; and  2d,  a point  where  two  or  more  branches  are 
tangent  to  each  other.  The  latter  are  sometimes  called 
points  of  osculation. 

As  each  branch  of  the  curve  has  its  tangent,  there  will 
be  at  a multiple  point  as  many  tangents,  and  therefore  as 

many  values  of  as  there  are  branches  which  meet  in 

this  point.  If  these  branches  are  all  tangent,  the  values  of 
dv 

will  be  equal.  At  a multiple  point  y will  have  but  one 

value,  while  at  points  near  it,  it  will  have  two  or  more 
values  for  each  value  of  x.  In  functions  of  a simple  form, 
such  a point  can  generally  be  determined  by  inspection. 
After  finding  a value  of  x for  which  y has  but  one  value, 
and  on  both  sides  of  which  it  has  two  or  more  values,  form 

— • If  this  has  unequal  values,  the  branches  of  the  curve 
dx 

intersect  at  this  point,  and  the  point  is  of  the  first  kind.  If 

has  but  one  value,  the  branches  are  tangent  to  each 
dx 

other  at  this  point,  and  the  point  is  of  the  second  kind. 

When  the  critical  points  are  not  readily  found  by  inspec- 
tion, we  proceed  as  follows : 

Let  f(x,  y)  — 0 

be  the  equation  of  the  locus  freed  from  radicals.  Then 


(1) 


EXAMPLES. 


197 


du 

dy  dx 

dx  ~ du  ’ 

dy 

and  as  differentiation  never  introduces  radicals  when  they 
do  not  exist  in  the  expression  differentiated,  the  value  of 

--  cannot  contain  radicals,  and  therefore  cannot  have  sev- 
dx 

eral  values,  unless  by  taking  the  form  jj- 

tt  i 0 du  du 

Hence  we  have  - or  — = 0,  and  7-  = 0,  from 

dx  0 dx  dy 

which  to  determine  critical  values  of  x and  y.  If  these 

dxo  du 

values  of  x and  y found  from  , = 0 and  — = 0 are  real 
J dx  dy 

and  satisfy  (1),  they  may  belong  to  a multiple  point.  If  y 

has  but  one  value  for  the  corresponding  value  of  x,  and  on 

both  sides  of  it  y has  two  or  more  real  values,  this  point  is  a 

multiple  point.  We  then  evaluate  ^ = jj,  and  if  there 

are  several  real  and  unequal  values  of  ~ , there  will  be  as 

dx 

many  intersecting  branches  of  the  curve  passing  through 
the  point  examined. 


EXAMPLES. 


1.  Determine  whether  the  curve  y 
a multiple  point. 

Here  y has  two  values  for  every 
positive  value  of  x > or  < a.  When 
x = 0 or  a,  y has  but  one  value,  i; 
hence  there  are  two  points  to  be  ex- 
amined. When  x < 0,  y is  imagi- 
nary ; hence  the  branches  do  not 
pass  through  the  point  (0,  l),  and 


= (x  — a)Vx  + i haa 


198 


EXAMPLES. 


therefore  it  is  not  a multiple  point.  When  x > or  < a,  y 
has  two  real  values,  and  therefore  (a,  l)  is  a double  point. 


dy 

dx 


± 


3x  — a 

%Vx 


— rfc  Va, 


when  x — a. 


Therefore  the  point  is  of  the  first  kind,  and  the  tangents 
to  the  curve  at  the  point  make  with  the  axis  of  x angles 
whose  tangents  are  -f-  V a and  — V a. 


2.  Examine  xi  -f-  2ax2y  — ay3  = 0 for  multiple  points. 


We  proceed  according  to  the  second  method,  as  all  the 
critical  points  in  this  example  are  not  easily  found  by  inspec- 
tion. 


du 

dx 


= 4 x ( x 2 + ay)  — 0 ; 


(1) 


du 

dy 


= a (2x2  — 3 y2)  — 0 ; 


dy  _ 4X3  -p  4 axy 
dx  3 ay2  — 2 ax2 

Solving  (1)  and  (2)  for  x and  y,  we  find 


(2) 

(3) 


lx  — 0\  lx  — -|-aA/6\  lx  = — 

V/  — 0/  \y  = -f  a)]  \y  = -fa  / 


Only  the  first  jrnir  will  satisfy  the  equa- 
tion of  the  curve,  and  therefore  the  ori- 
gin is  the  only  point  to  be  examined. 

Evaluating  ~ in  (3)  for  x — 0 and 
° dx  v ’ 

y = 0,  and  representing  (~  by  p,  and  ^ 
by  p\  for  shortness,  we  have 


CUSPS. 


199 


dy  lx3  -f-  4 axy  0 

dx  ^ day2  — 2 ax2  0 ’ 

_ 12a;2  + 4 ay  + 4 axp  0 

Gayp  - 4 ax  0 ’ 

_ 24a;  -f  Sap  + 4=axp' 

Gap2  + 6 ayp'  — 4 a 
Sap 

Gap2  — 4a  ’ 


p (Gap2  — - 4a)  = Sap ; 


> = l = * 


+ V%,  or  — a/2. 


Hence  the  origin  is  a triple  point,  the  branches  being  in- 
clined to  the  axis  of  x at  the  angles  0,  tan-1  (a/2),  and 
tan-1( — a/2),  respectively,  as  in  the  figure.  (See  Courte- 
nay’s Calculus,  p.  191 ; or  Young’s  Calculus,  p.  151.) 

3.  Examine  y2  — a:2  (1  — a’2)  = 0 for  multiple  points. 
Ans.  There  is  a double  point  at  the  origin,  the  branches 

being  inclined  to  the  axis  of  a?  at  angles  of  45°  and  135° 
respectively. 

4.  Show  that  ay3 — x3y — aa?  = 0 has  no  multiple  points. 


111.  Cusps. — A cusp  is  a point  of  a curve  at  which  two 
branches  meet  a common  tangent,  and 
stop  at  that  point.  If  the  two  branches 
lie  on  opposite  sides  of  the  common  tan- 
gent, the  cusp  is  said  to  be  of  the  first 
species ; if  on  the  same  side,  the  cusp  is 
said  to  be  of  the  second  species. 

Since  a cusp  is  really  a multiple  point 
of  the  second  kind,  the  only  difference 
being  that  the  branches  stop  at  the  point, 
instead  of  running  through  it,  we  exam- 
ine for  cusps  as  we  do  for  multiple  points ; and  to  distin- 


200 


CUSPS. 


guish  a cusp  from  an  ordinary  multiple  point,  we  trace  the 
curve  in  the  vicinity  of  the  point  and  see  if  y is  real  on  one 
side  and  imaginary  on  the  other.  To  ascertain  the  kind  of 
cusp,  we  compare  the  ordinates  of  the  curve,  near  the  point, 
with  the  coi'responding  ordinate  of  the  tangent ; or  ascertain 
the  direction  of  curvature  by  means  of  the  second  derivative 
In  the  particular  case  in  which  the  common  tangent  to 
the  two  branches  is  perpendicular  to  the  axis  of  x,  it  is  best 
to  consider  y as  the  independent  variable,  and  find  the 
rJr 

values  of  , etc. 

dy 


EXAMPLES. 

1.  Examine  y = x2  ± x?  for  cusps. 

We  see  that  when  x — 0,  y has  but  one  value,  0;  when 
x < 0,  y is  imaginary  ; and  when  x > 0,  y has  two  real 
values;  hence,  (0,  0)  is  the  point  to  be  examined. 

— 2x  + = 0,  when  x — 0 ; hence  the  axis  of  x is 

dx  4 

a common  tangent  to  both  branches,  and 
there  is  a cusp  at  the  origin. 
cPy  1 

= 2±^-x-  is  positive  when  x = 0 ; 

hence  the  cusp  is  of  the  second  kind. 

(fill 

The  value  of  shows  that  the  upper 

branch  is  always  concave  upward,  while  the  lower  branch 
has  a point  of  inflexion,  when  x = ; from  the  origin  to 

the  point  of  inflexion  this  branch  is  concave  upward,  after 
which  it  is  concave  downward. 

The  value  of  ^ shows  that  the  branch  is  horizontal 
dx 

when  x — \\.  From  y — x2 — x%  we  find  that  the  lower 
branch  cuts  the  axis  of  x at  x 1.  The  shape  of  the  curve 
is  given  in  Fig.  28. 


CONJUGATE  POINTS. 


201 


2.  Examine  ( y — i)2  — (x  — a)3  for  cusps. 

Ans.  The  point  (a,  i)  is  a cusp  of  the  first  kind. 
3 Examine  cy 2 = x3  for  cusps. 

The  origin  is  a cusp  of  the  first  kind. 


112.  Conjugate  Points. — A conjugate  point  is  an  iso 
lated  point  whose  co-ordinates  satisfy  the  equation  of  the 
curve,  while  the  point  itself  is  entirely  detached  from  every 
other  point  of  the  curve. 

For  example,  in  the  equation  y — (a  + x)  \/x,  if  x is 
negative,  y is,  in  general,  imaginary  but  for  the  particular 
value  x = — a,  y = 0.  Hence,  P is  a 
point  in  the  curve,  and  it  is  entirely 
detached  from  all  others.  When  x = 0, 
y = 0,  which  shows  that  the  curve 
passes  through  the  origin.  For  positive 
values  of  x,  there  will  be  two  real  values 
of  y,  numerically  equal,  with  opposite 
signs.  Hence,  the  curve  has  two  infinite  branches  on  the 
right,  which  are  symmetrical  writ,h  respect  to  the  axis  of  x. 

If  the  first  derivative  becomes  imaginary  for  any  real 
values  of  x and  y,  the  corresponding  point  will  be  conjugate, 
as  the  curve  will  then  have  no  direction.  It  does  not  fol- 


low, however,  that  at  a conjugate  point  will  be  imagi- 
nary; for,  if  the  curve  y =f(x)  have  a conjugate  point  at 
(x,  y),  from  the  definition  of  a conjugate  point,  we  shall 
have 


f{x±_  h ) = an  imaginary  quantity.  But 


f(x  ±li)  — y ± 


dy  li 
dx  1 


d2y  h2  d3y  hs 
— — — -4-  — — — -I-  oto 
dx2  2 -dtf6  ^ ’ 


therefore,  if  either  one  of  the  derivatives  is  imaginary,  the 
first  member  is  imaginary;  hence,  at  a conjugate  point 
some  one  or  more  of  the  derivatives  is  imaginary. 

Since  at  a conjugate  point  some  of  the  derivatives  are 
imaginary,  let  the  nth  derivative  be  the  first  that  is  imagi- 


202 


EXAMPLES. 


nary.  Suppose  the  equation  of  the  curve  to  be  freed  from 
radicals,  and  denoted  by  u = / (x,  y ) = 0.  Take  the  vP1 
derived  equation  (Art.  88,  Sch.) ; we  have 


du  dny 
dy  dxn 


dnu 

+ SF  = °- 


where  the  terms  omitted  contain  derivatives  of  u with  re 
spect  to  x and  y,  and  derivatives  of  y with  respect  to  x,  of 

cLu 

lower  orders  than  the  nth.  If,  then,  -7-  be  not  0,  the  value 

dy 

dny 

of  ~n  obtained  from  the  derived  equation  will  be  real, 

du 

which  is  contrary  to  the  hypothesis;  hence,  -—  = 0 is  a 

necessary  condition  for  the  existence  of  a conjugate  point. 
But 

du  dudy_ 
dx  + dy  dx  — ’ 


therefore,  since 


^ = 0,  we  must  have  ~ — 0. 
dy  dx 


Hence,  at 


a conjugate  point  we  must  have  ^ = 0,  and  ~ = 0. 


Rem. — Owing  to  the  labor  of  finding  the  higher  derivatives,  it  is 
usually  better,  if  the  first  derivative  does  not  become  imaginary,  to 
substitute  successively  a + h and  a — h for  x,  in  the  equation  of  the 
curve,  where  a is  the  value  of  x to  be  tested,  and  h is  very  small.  If 
both  values  of  y prove  imaginary,  the  point  is  a conjugate  point. 


EXAMPLES. 

1.  Examine  ay 2 — x?  + 4 ax2  — 5 a2x  + 2 a3  = 0 for  con 
jugate  points. 

(^U-  = — 3a:2  + 8 ax  — 5 a2  = 0.  (1) 

w 


I = ^ = °- 


SHOOTING  POINTS. — STOP  POINTS. 


203 


Solving  (1)  and  (2),  we  get 


Only  the  first  pair  of  values  satisfies  the  equation  of  the 
curve,  and  hence  the  point  ( a , 0)  is  to  be  examined. 


dy  3z2  — 8 ax  -f-  5a2  6x  — 8 a 

dx  — ^ — 'lay  ~~  lap 


therefore, 


1 , (x  — a\ 

= , when  ( ); 

p \y  = o/ 

P‘=-1>  P = ± V- 1 = %■ 


This  result  being  imaginary,  the  point  (a,  0)  is  a conju- 
gate point. 

2.  Show  that  re4  — ax2y  — axy2  + a2y 2 = 0 has  a conju- 
gate point  at  the  origin. 

3.  Examine  ( c2y  — a;3)2  = (x  — a)5  (x  — b)6  for  conjugate 
points,  in  which  a > b. 

The  point  (b,  — ^ is  a conjugate  point. 

The  first  and  second  derivatives  are  real  in  this  example  ; hence  the 
better  method  of  solving  it  will  be  to  proceed  according  to  the  Remark 
above  given 

113.  Shooting  Points  are  points  at  which  two  or  more 
branches  of  a curve  terminate,  without  having  a common 
tangent. 

Stop  Points  are  points  in  which  a single  branch  of  a 
curve  suddenly  stops. 

These  two  classes  of  singular  points  but  rarely  occur,  and 
never  in  curves  whose  equations  are  of  an  algebraic  form. 


204 


EXAMPLES. 


EXAMPLES. 


1.  Examine  y = 


l + ez 


for  shooting  points. 


Here 


dy 

dx 


1 + e* 


1 + 


(l  +*ir 


If  x is  + and  small,  y is  + ; if 
x is  — and  small,  y is  — . When 
x is  + and  approaches  0,  y — 0, 


and  — 0 ; when  a;  is  — 
dx 

proaches  0,  y = 0,  and 


and  ap- 

fy  _ i 
dx~  ‘ 


Fig.  30. 


Hence,  at  the  origin  there  is  a shooting  point,  one  branch 
having  the  axis  of  x as  its  tangent,  and  the  other  inclined 
to  the  axis  of  x at  an  angle  of  45°.  (See  Serret’s  Calcul 
Diflerentiel  et  Integral,  p.  267.) 

2.  Examine  y — x log  x. 

When  a;  is  +,  y has  one  real  value ; when  x = 0,  y — 0 ; 
when  x < 0,  y is  imaginary  ; hence  there  is  a stop  point  at 
the  origin. 

3.  Examine  y = x tan-1  ^ 


x 

= tan-  1 
dx  x 


x 


If 


:2  + l 

a;  = + 0 or  — 0,  y = 0 ; 

dy 7z  ' 7T 

dx  ~ 2 °r  — 2‘ 


Hence  the  origin  is  a shooting  point,  the  tangent  being 
inclined  to  the  axis  of  x at  angles  tan-1  (1.5708)  and 
tan-1  (-  1.5708). 

4.  Show  that  y = e x has  a stop  point  at  the  origin. 


TRACING  curves. 


205 


114.  Tracing  Curves. — We  shall  conclude  this  chap- 
ter by  a brief  statement  of  the  mode  of  tracing  curves  by 
means  of  their  equations. 

The  usual  method  of  tracing  curves  consists  in  assigning 
a series  of  different  values  to  one  of  the  variables,  and  cal- 
culating the  corresponding  series  of  values  of  the  other,  thus 
determining  a definite  number  of  points  on  the  curve.  By 
drawing  a curve  or  curves  through  these  points,  we  are 
enabled  to  form  a tolerably  accurate  idea  of  the  shape  of  the 
curve.  (See  Anal.  Geometry,  Art.  21.) 

In  the  present  Article  we  shall  indicate  briefly  the  man- 
ner of  finding  the  general  form  of  the  curve,  especially  at 
such  points  as  present  any  peculiarity,  so  that  the  mind  can 
conceive  the  locus,  or  that  it  may  be  sketched  without 
going  through  the  details  of  substituting  a series  of  values, 
as  was  referred  to  above. 

To  trace  a curve  from  its  equation,  the  following  steps 
will  be  found  useful : 

( 1 .)  If  it  be  possible,  solve  it  with  respect  to  one  of  its 
variables,  y for  example,  and  observe  whether  the  curve  is 
symmetrical  with  respect  to  either  axis. 

(2.)  Find  the  points  in  which  the  curve  cuts  the  axes, 
also  the  limits  and  infinite  branches. 

(3.)  Find  the  positions  of  the  asymptotes,  if  any,  and  at 
which  side  of  an  asymptote  the  corresponding  branches  lie. 

(f)  Find  the  value  of  the  first  derivative,  and  thence 
deduce  the  maximum  and  minimum  points  of  the  curve,  the 
angles  at  which  the  curve  cuts  the  axes,  and  the  multiple 
points,  if  any. 

(5.)  Find  the  value  of  the  second  derivative,  and  thence 
the  direction  of  the  curvature  of  the  different  branches,  and 
the  points  of  inflexion,  if  any. 

(6.)  Determine  the  existence  and  nature  of  the  singular 
points  by  the  usual  rules. 


206 


EXAMPLES. 


EXAMPLES. 


1.  Trace  the  curve  y ■=  =• 

a 1 + x2 


When  x = 0,  y = 0 ; the  curve  passes  through  the 
origin. 

For  all  positive  values  of  x,  y is  positive  ; and  when 
x = co  , y — 0.  For  negative  values  of  x,  y is  negative,  and 
when  x — — oo , y — 0 ; hence  the  curve  has  two  infinite 
branches,  one  in  the  first  angle  and  one  in  the  third,  and  the 
axis  of  x is  an  asymptote  to  both  branches. 

dy  _ 1 — x1  d2y  2x  (x2  — 3) 

dx  ~ (1  + x2)2’  dx2  ~~  (1  -f  x2)3 


When  z = ± 1,  g = 0; 


.\  there  is  a maximum  ordinate 


at  x = + 1,  and  a minimum  ordinate  at 
which  points  y = 4 and  — respectively. 


x—  — 1,  at 


0 ^-1- 
dx  ~ ’ 


When  x = 0,  ^r-  = 1 ; .'.  the  curve  cuts  the  axis  of  x at 
an  angle  of  45°. 

Putting  the  second  deriva- 
tive equal  to  0,  we  get  x — 0 
or  ± V3.  Therefore,  there 
are  points  of  inflexion  at  (0,0) 
and  at  x — + \/3  and  — a/3, 
for  which  we  have  y = £\/3, 

— Ja/3.  From  x — — V3 
to  x — -f  a/3,  the  curve  is  concave  towards  the  axis  of  x, 
and  beyond  them  it  is  convex. 

From  this  investigation  the  curve  is  readily  constructed, 
and  has  the  form  given  in  the  figure. 


2.  Trace  the  curve  y3  — 2 ax2  — z3. 

y — x$  (2a  — x)i  ; 


EXAMPLES. 


207 


dy  _ 4 ax  — 3 x2  _ d2?/  — 8a2 

rfa:  _ 3«/2  ’ 5S2  ~ 9a:t  (2 a - xf 

When  x = 0 or  2cr,  y — 0 ; .\  the  curve  cuts  the  axis  of 
x at  the  origin  and  at  x = 2a. 

To  find  the  equation  of  the  asymptote,  we  have 


therefore,  y = — x + f a is  the  equation  of  the  asymptote, 
and  as  the  next  term  of  the  expression  is  positive,  the  curve 
lies  above  the  asymptote. 

Evaluating  the  first  derivative  for  x = 0,  y = 0,  we  have 


dy  4=ax  — 3a;2 

dx  ~ 3 y* 


4 a — Qx 

a dy  '' 


when  x — y — 0 ; 


= ± oo , when  «/  = ([. 

Hence,  at  the  origin  there  are 
two  branches  of  the  curve  tangent 
to  the  axis  of  y,  and  the  value  of 
dy 

~ shows  that  if  y be  negative  as  it 

approaches  0,  ^ will  be  imaginary; 

and  hence  the  origin  is  a cusp  of 
the  first  species. 

When  a;  = 4 a,  — =0  \ :.  there  is  a maximum  ordinate 
, d dx 
at  x — ■§«. 

dv  Aicd" 

When  x = 2a.  -f-  = = — oo ; .-.the  curve  cuts 

dx  0 

the  axis  of  x,  at  the  point  x = 2a,  at  right  angles. 


208 


EXAMPLES. 


Futting  the  second  derivative  equal  to  oo,  we  get  x = 2a. 
When  x < 2a,  the  second  derivative  is  — , and  when  > 2 a 
it  is  + ; hence  the  left  branch  is  everywhere  concave  down- 
ward, and  the  right  branch  is  concave  downward  from  x = 0 
to  x — 2a.  At  this  last  point  it  cuts  the  axis  of  x at  right 
angles,  and  changes  its  curvature  to  concave  upward;  the 
two  branches  touch  the  asymptote  at  x = + qo  and  — oo 
respectively,  i.  e.,  they  have  a common  asymptote. 

In  the  figure,  OA  = 2a,  OB  ■§■«,  OC  = \a. 


3.  Trace  the  curve  y 

« 

1 1 
II 

o 

II 

& 

y = o. 

z<a; 

y is  positive. 

x — a; 

y = «• 

x > a < 2a  ; 

y is  negative. 

e 

II 

y = o. 

x > 2a’, 

y is  positive. 

x — co ; 

y = 00  • 

When  x is  — -,  y is  always  negative. 

To  find  the  asymptote,  we  have 

x(1-2i)il  + l + etc) 

= x(l  — - — ‘~x  ~ etc. ) = x — a — etc. 

y = x — a is  the  equation  of  the  asymptote. 

Hence,  take  OB  — a — OD,  and  the  line  BD  produced 
is  the  asymptote;  also  take  OC  = 2a.  Then,  since  y = 0, 
kcth  when  x — 0 and  x = 2a,  the  curve  cuts  the  axis  of  x 


EXAMPLES. 


209 


at  0 and  C.  Between  0 and  B,  the  curve  is  above  the  axis ; 
at  B the  ordinate  is  infinite  ; from  B to  C,  the  curve  is 
below;  from  C to  infinity,  it  is  above  OX.  Also,  if  x is 
negative,  y is  negative ; therefore  the  branch  on  the  left  of 
0 is  entirely  below  the  axis. 


Also, 


dy  x2  — 2 ax  + 2a2 

dx  — (x  — a)2 


Let  x — a ; = oo ; and  the  infinite  ordinate  at  the 

distance  a to  the  right  of  the  origin  is  an  asymptote. 

If  x — 0,  ^'|  = 2;  if  x = 2a,  = 2‘,  i.  e.,  the  curve 

cuts  the  axis  of  x at  the  origin  and  the  distance  2 a to  the 

right,  at  the  same  angle,  tan  _1  (2). 

If  x 2 — 2 ax  + 2 a2  or  (x  — a)2  + a2  = 0,  £ is  impossible; 
hence  there  is  no  maximum  or  minimum  ordinate. 


Again, 


d?y 2 [x  — a)2  — 2 [(a;  — a)2  + a2] 

dx2  — (x  — a)3 

— 2a2 

— (x  — a)3  ’ 


is 


+ if  x < a,  and  is  — ii  x a. 


But  x < a,  y is  + ; and  x > a < 2a,  y is  — ; and 
x > 2a,  y is  + ; therefore,  from  0 to  B,  and  B to  C,  the 
curve  is  convex,  and  from  C to  infinity,  it  is  concave  to  the 
axis  of  x. 


If  x be 


d2y 

dx* 


2 a2 


is  +,  but  y is  — ; therefore 


(x  + a)3 

the  branch  from  the  origin  to  the  left  is  concave  to  the  axis 
of  x.  (See  Hall’s  Calculus,  pp.  182,  183.) 


4.  Trace  the  curve  y2  = a2x 3. 

The  curve  passes  through  the  origin  ; is  symmetrical 
with  respect  to  the  axis  of  x\  has  a cusp  of  the  first  kind  at 


210 


TRACING  POLAR  CURVES. 


the  origin;  both  branches  are  tangent  to  the  axis  of  x ; are 
convex  towards  it;  are  infinite  in  the  direction  of  positive 
abscissas,  and  the  curve  has  no  asymptote  or  point  of  in- 
flexion. 

115.  On  Tracing  Polar  Curves. — Write  the  equation, 
if  possible,  in  the  form  r =/ (0)  ; give  to  0 such  values  as 
to  make  r easily  found,  as  for  example,  0,  r,  n,  -|tt,  etc. 
civ 

Putting  -jg  = 0,  we  find  the  values  of  0 for  which  r is  a 

maximum  or  minimum,  i.  e.,  where  the  radius  vector  is 
perpendicular  to  the  curve. 

Find  the  asymptotes  and  direction  of  curvature,  and 
points  of  inflexion.  After  this  there  will  generally  be  but 
little  difficulty  in  finding  the  form  of  the  curve. 

EXAMPLES. 

1.  Trace  the  lituus  r — 

05 

When  0 = 0,  r = oo  ; when  0 = 1 (=  57°.3),*  r = ± a ; 
when  0 = 2 (=  114°.6),  r = Si  a ; when  0 = 3,  r = 
± .58 a,  etc. ; when  0 = oo , r = 0. 

dv  7’ ^ (Hy* 

-=  = — and  when  — = 0,  r = 0 ; hence,  r and  6 
dv  Za4  do 

are  decreasing  functions  of  each  other  throughout  all  their 
values;!  and  the  curve  starts  from  infinity,  when  0 = 0, 
and  makes  an  infinite  number  of  revolutions  around  the 
pole,  cutting  every  radius-vector  at  an  oblique  angle,  and 
reaching  the  pole  only  when  0 = oo . 
dd 

The  subtangent  r2  — = — — = 0,  when  r = oo  ; hence 
dr  r ’ 

the  initial  line  is  an  asymptote  (Art.  105). 


* The  unit  angle  is  that  whose  arc  is  equal  to  the  radius,  and  is  about  5?'. 29578 
t If  we  consider  alone  the  branch  generated  by  the  positive  radius-vector. 


EXAMPLES. 


211 


dj)  _ 2a  (4 a r*)  ^gee  Art.  Ex.  3)  ; hence  there  is 

df  (4«4  + r4)l 

a point  of  inflexion  at  r = aV 2 ; from  r — 0 to  r = aV'2 
the  curve  is  concave  toward  the  pole,  and  from  r — aV 2 to 
r — oo  it  is  convex. 

2.  Trace  the  curve  r = a sin  3d. 


r = 0,  when  0=0,  60°,  120°,  180°,  240°,  and  300°. 
When  d = 2tt,  or  upwards,  the  same  series  of  values  recur. 

If  e = 30°,  90°,  150°,  210°,  270°,  and  330°,  r = a,  — a, 
a,  — a,  a,  and  — a,  successively. 
dv 

—~  = 3a  cos  3d,  showing  that  r begins  at  0 when  d = 0, 


increases  till  it  is  a when  d = 30°,  diminishes  to  0 as  d 
passes  from  30°  to  60°,  continues  to  diminish  and  becomes 
— a when  6 becomes  90°,  and  so  on. 
dp  _ 18  a2r  — 8r3 
dr 


which  shows  that 


(9«2  _ 8r2)f 

the  curve  is  always  concave  towards  the 
pole.  There  is  no  asymptote,  as  r is 
never  oo . 

Hence  the  curve  consists  of  three 
equal  loops  arranged  symmetrically 
around  the  pole,  each  loop  being  traced  twice  in  each  revo^ 
lution  of  r.  A little  consideration  will  show  that  the  form 
of  the  curve  is  that  given  in  the  figure.  (See  Gregory’s 
Examples,  p.  185  ; also  Price’s  Calculus,  Vol.  I,  p.  427.) 


3.  Trace  the  Chordel  r = a cosec  ( — 


If  d = 0,  tin,  2nn,  3 nn,  4/z~,  5nn,  etc.,  successively, 
r = oo  , a,  oo , — a,  — <x> , a,  etc. 


dr  a 6 6 a „ 6 . 

M = - Tn  C0SeC  ii,  Cot  = Wi  C0Se°  & 1 


:(-cos!)’ 

which  is  negative  from  d = 0 to  d = nn,  positive  from 


212 


EXAMPLES. 


0 = nn  to  0 = 3 nn,  negative  from  0 3nn  to  0 = onn,  etc. 

Hence  we  see  that  r begins  at  co  when  0 = 0;  diminishes 
till  it  becomes  a when  0 = nn-,  increases  as  0 passes  from 
nn  to  2nn  ; becomes  c c when  0 = 2nn ; when  0 passes  2nn, 
r changes  from  + o>  to  — oo ; when  6 increases  from  2nn 
to  ?>nn,  r increases  from  — oo  to  — a;  when  6 increases 
from  3nn  to  4 nn,  r diminishes  from  — a to  — oo ; when 
6 passes  4 nn,  r changes  from  — oo  to  + oo . When  6 in- 
creases beyond  4~,  the  same  values  of  r recur,  showing  that 
the  curve  is  complete. 


dr  a „ 0 / 0\a  - a q 

= — cosec2  =- 1 — cos  =- 1 = 0 gives  0 = nn,  3nn, 
dd  2 n 2 n \ 2 n! 

bnn,  etc. ; i.  e.,  the  radius-vector  is  a minimum  at  0 = nn, 
3 nn,  bnn,  etc. 

The  subtangent  = r2^  = 

° dr  0 

C0S  2n 

= — 2 na  when  0 = 0; 
and  = -(-  2 na  when  0 = 2nn  ; 


EXAMPLES. 


213 


therefore  the  curve  has  two  asymptotes  parallel  to  the  initial 
line,  at  the  distances  ± %na  from  the  pole. 

r2  2 anr 

P~((W  2\i  ~ ua2n2  - a2  + r2)i  ’ 

\dd2  ^ ) 

dp  2a3n  (4«2 — 1) 

dr  ~ [a2  {in2  - 1)  + r2]i  ’ 

.\  the  curve  is  always  concave  towards  the  pole. 

Thus  it  appears  that  while  6 is  increasing  from  0 to  2 nn, 
the  positive  end  of  the  radius-vector  traces  the  branch 
drawn  in  Fig.  34;  and  while  6 increases  from  2nn  to  4 mr, 
the  negative  end  of  the  radius-vector  traces  a second  branch 
(not  drawn),  the  two  branches  being  symmetrical  with 
respect  to  the  vertical  line  through  the  pole  0. 


EXAMPLES. 

1.  Find  the  direction  of  curvature  of  the  Witch  of  Agnesi 

x2y  — 4 a2  (2 a — y ). 

The  curve  is  concave  downward  for  all  values  of  y between 
2 a and  fa,  and  convex  for  all  values  of  y between  \a  and  0. 

2.  Find  the  direction  of  curvature  of  y = b + (x  — a)3. 

Convex  towards  the  axis  of  x from  x > a to  x = qo  ; and 
from  x = a — $ to  x — — go  ; concave  towards  the  axis 
of  x from  x < a to  x — a — b^. 

3.  Examine  y — {a  — a;)^  + ax  for  points  of  inflexion. 

There  is  a point  of  inflexion  at  x — a. 

4.  Examine  y — x - f-  3Qx2  — 2a;3  — a^  for  points  of  in- 
flexion. Points  of  inflexion  at  x = 2,  x = — 3. 


214 


EXAMPLES. 


5.  Find  the  co-ordinates  of  the  point  of  inflexion  of  the 
curve 

x?  ( a 2 — x?) 


y = 


ad 2 


*=±vl;  y = Aa' 


6.  Examine  r = s z ior  points  of  inflexion. 

cr  — 1 

dr2  4?-  ( r — a )3 


Here 


dd2 
P = 


.*.  etc. 


(4?4  — 12  a?-3  + 13«2r2  — 4 asr)^ 

There  are  points  of  inflexion  at  r = fa  and  r = \a. 


7.  Examine  y2  — {x  — Vfx  for  multiple  points. 

There  is  a multiple  point  at  x — 1. 


8.  Examine  y 2 = 


x2  ( a 2 — z2) 
a2  + a2 


for  multiple  points. 


There  is  a multiple  point  at  the  origin,  and  the  curve  is 
composed  of  two  loops,  one  on  the  right  and  the  other  on 
the  left  of  the  origin,  the  tangents  bisecting  the  angles  be- 
tween the  axes  of  co-ordinates. 


9.  Show  that  x*  + x2y2  — 6 ax*y  + a2y 2 = 0 has  a multiple 
point  of  the  second  kind  at  the  origin. 

10.  Show  that  y = a + x + bx2  i cxi  has  a cusp  of  the 
second  kind  at  the  point  (0,  a),  and  that  the  equation  of  the 
tangent  at  the  cusp  is  y = x + a. 

11.  Show  that  y3  = ax 2 + a'3  has  a cusp  of  the  first  kind 
at  the  origin. 

12.  Show  that  ay2  — a3  + bx2  — 0 has  a conjugate  point 

4 i 

at  the  origin,  and  a point  of  inflexion  at  x = — • 

O 


EXAMPLES. 


215 


13.  Trace  the  curve  ys  — a3  — x?. 

The  curve  cuts  the  axes  at  (a,  0)  and  (0,  a). 

It  has  an  asymptote  which  passes  through  the  origin. 

The  points  where  the  curve  cuts  the  axes  are  points  ol 
inflexion. 

14.  Trace  the  curve  y = ax2  d:  Vbx  sin  x. 

For  every  positive  value  of  x there  are 
two  values  of  y,  and  therefore  two  \ 
points,  except  when  sin  x = 0,  in  which  \ 
case  the  two  points  reduce  to  one.  N. 

These  points  form  a series  of  loops  like  pig.  35, 

the  links  of  a chain,  and  have  for  a 
diametral  curve  the  parabola  y — ax1,  from  which,  when  x 
is  positive,  the  loops  recede  and  approach,  meeting  the 
parabola  whenever  x — 0 or  tt,  or  any  multiple  of  rr.  But 
when  x is  negative,  y is  imaginary  except  when  sin  x — 0, 
in  which  case  y — ax?,  so  that  on  the  negative  side  there  is 
an  infinite  number  of  conjugate  points,  each  one  on  the 
parabola  opposite  a double  point  of  the  curve.  (See  De 
Morgan’s  Cal.,  p.  382;  also,  Price’s  Cal.,  Yol.  I,  p.  396.) 


CHAPTER  XI. 


RADIUS  OF  CURVATURE,  EVOLUTES  AND  INVO 
LUTES,  ENVELOPES. 

116.  Curvature.  — The  curvature  of  a curve  is  its  rate 
of  deviation  from  a tangent , and  is  measured  by  the  external 
angle  between  the  tangents  at  the  extremities  of  an  indefi- 
nitely small  arc  ; that  is,  by  the  angle  between  any  infini- 
tesimal element  and  the  prolongation  of  the  preceding 
element.  This  angle  is  called  the  angle  of  contmgence  of 
the  arc.  Of  two  curves,  that  which  departs  most  rapidly 
from  its  tangent  has  the  greatest  curvature.  In  the  same 
or  in  equal  circles,  the  curvature  is  the  same  at  every 
point ; but  in  unequal  circles,  the  greater  the  radius  the 
less  the  curvature ; that  is,  in  different  circles  the  curvature 
varies  inversely  as  their  radii. 

Whatever  be  the  curvature  at 
any  point  of  a plane  curve,  it  is 
clear  that  a circle  may  be  found 
which  has  the  same  curvature  as 
the  curve  at  the  given  point,  and 
this  circle  can  be  placed  tangent 
to  the  curve  at  that  point,  with 
its  radius  coinciding  in  direction 
with  the  normal  to  the  curve  at 
the  same  point.  This  circle  is  called  the  osculating  circle, 
or  the  circle  of  curvature  of  that  point  of  the  curve.  The 
radius  of  curvature  is  the  radius  of  the  osculating  circle. 
The  centre  of  curvature  is  the  centre  of  the  osculating  circle. 

For  example,  let  ABA  B be  an  ellipse.  If  different 
circles  be  passed  through  B with  their  centres  on  BB , it  is 


B 


ORDER  OF  CONTACT  OF  CURVES. 


217 


clear  that  they  will  coincide  with  the  ellipse  in  very  differ- 
ent degrees,  some  falling  within  and  others  without.  Now, 
that  one  which  coincides  with  the  ellipse  the  most  nearly 
of  all  of  them,  as  in  this  case  MN,  is  the  osculating  circle 
of  the  ellipse  at  B,  and  is  entirely  exterior  to  the  ellipse. 
The  osculating  circle  at  A or  A',  is  entirely  within  the 
ellipse  ; while  at  any  other  point,  as  P,  it  cuts  the  ellipse, 
as  will  be  shown  hereafter. 

117.  Order  of  Contact  of 
Curves.  — Let  y = f (x)  and  y 
— 0 ( x ) be  the  equations  of  the 
two  curves,  AB  and  ab,  referred  to 
the  axes  OX  and  OY.  Giving  to 
x an  infinitesimal  increment  h,  and 
expanding  by  Taylor’s  theorem,  we 
have, 

Vi  = + h)  =f(x)  + f (-L)  ^ + f"  (*)  \ 

7? 3 

+ /"'  0)  2T3  + etc- 

Ji 2 

y,  — 0 (x  + h)  = 0 (x)  + 0'  (x)  h + 0"  (x)  - 

+ </>'"  (*)  ^7-3  + etc.  (2) 


(1) 


Now  if,  when  x = a = OM,  we  have  / (a)  =0  (a),  the 
two  curves  intersect  at  P,  i.  e.,  have  one  point  in  common. 
If  in  addition  we  have  /'  (a)  — 0'  (a),  the  curves  have  a 
common  tangent  at  P,  i.  e.,  have  two  consecutive  points  in 
common  ; in  this  case  they  are  said  to  have  a contact  of  the 
first  order.  If  also  we  have,  not  only/ (a)  = 0 (a)  and/'  (a) 
= 0'  ( a ),  but  /"  (a)  = 0"  (a) ; i.  e.,  in  passing  along  one  of 

d?y 
dx2 

ture),  remains  the  same  in  both  curves,  and  the  new  point 
10 


the  curves  to  the  next  consecutive  point,  (i.e.,  the  curva- 


218 


CONTACT  OF  THE  SECOND  ORDER. 


is  also  a point  of  the  second  curve ; i.  e.,  the  curves  have 
three  consecutive  points  in  common  ; in  this  case  the  curves 
are  said  to  have  a contact  of  the  second  order.  If  f (a) 
= <P  («),  f (a)  = <P'  {a),  f"  (a)  = <p"  (a),  f"  (a)  = " (a), 

the  contact  is  of  the  third  order,  and  so  on.  It  is  plain 
that  the  higher  the  order  of  contact,  the  more  nearly  do 
the  curves  agree;  if  every  term  in  (1)  is  equal  to  the  cor- 
responding term  in  (2),  then  yx  — y2,  and  the  two  curves 
become  coincident. 

118.  The  Order  of  Contact  depends  on  the  num- 
ber of  Arbitrary  Constants. — In  order  that  a curve  may 
have  contact  of  the  nth  order  with  a given  curve,  it  follows 
from  Art.  117  that  n - f 1 equations  must  be  satisfied. 
Hence,  if  the  equation  to  a species  of  curve  contains  n -\-  1 
constants,  we  may  by  giving  suitable  values  to  those  con- 
stants, find  the  particular  curve  of  the  species  that  has 
contact  of  the  nth  order  with  a given  curve  at  a given  point. 
For  example,  the  general  equation  of  the  right  line  has  two 
constants,  and  hence  two  conditions  can  be  formed,  f ( x ) 
= (p  (a;)  and /'  (*)  <p'  ( x ),  from  which  the  values  of  the 

constants  may  be  determined  so  as  to  find  the  particular 
right  line  which  has  contact  of  the  first  order  with  a given 
curve  at  a given  point.  In  general,  the  right  line  cannot 
have  contact  of  a higher  order  than  the  first. 

Contact  of  the  second  order  requires  three  conditions, 
/ (x)  = <p  ( x ),  /'  (x)  = f ( x ),  and  f"  (x)  = <p"  ( x ),  and 
hence  in  order  that  a curve  may  have  contact  of  the  second 
order  with  a given  curve,  its  equation  must  contain  three 
constants,  and  so  on.  The  general  equation  of  the  circle 
has  three  constants  ; hence,  at  any  point  of  a curve  a circle 
may  be  found  which  has  contact  of  the  second  order  with 
the  curve  at  that  point ; this  circle  is  called  the  osculating 
circle  or  circle  of  curvature  of  that  point ; in  general, 
the  circle  cannot  have  contact  of  a higher  order  than 
the  second.  The  parabola  can  have  contact  of  the 


RADIUS  OF  CURVATURE. 


219 


third  order,  and  the  ellipse  and  hyperbola  of  the 
fourth. 

In  this  discussion  we  have  assumed  that  the  given  curve  is  of  such 
nature  as  to  allow  of  any  order  of  coutact.  Of  course  the  order  of 
contact  is  limited  as  much  by  one  of  the  curves  as  by  the  other.  For 
example,  if  the  given  curve  were  a right  line  and  the  other  a circle, 
the  contact  could  not  in  general  be  above  the  first  orier,  although  the 
circle  may  have  a contact  of  the  second  order  with  curves  whose 
equations  have  at  least  three  constants.  Also,  we  have  used  the 
phrase  in  general,  since  exceptions  occur  at  particular  points,  some  of 
which  will  be  noticed  hereafter. 

119.  To  find  the  radius  of  curvature  of  a given 
curve  at  a given  point,  and  the  co-ordinates  of  the 
centre  of  curvature. 

Let  the  equation  of  the  given  curve  be 


it  is  required  to  determine  the  values  of  m,  n,  and  r. 

Since  (2)  has  three  arbitrary  constants,  we  may  impose 
three  conditions,  and  determine  the  values  of  these  con- 
stants that  fulfil  them,  and  the  contact  will  be  of  the 
second  order  (Art.  118). 

From  (2),  by  differentiating  twice,  we  have. 


If  (2)  is  the  circle  of  curvature  at  the  point  ( x , y)  of  (1), 
»ve  must  have, 


V = / 00, 

and  that  of  the  required  circle  be 

(x’  — m )2  {y1  — n)2  = r2 ; 


(1) 


(3) 


(4) 


x'  = x,  y'  — y, 

dy'  dy  cPy'  d2y 

dx  — dx’  dx'2  dx2 


220 


RADIUS  OF  CURVATURE. 


Substituting  these  values  in  (2),  (3),  and  (4),  we  have, 


(x  — my  + (y  — n )2  = r2 ; 
x — m + (y  — n)<^  = 0 

1 + clt  , (v  _ n)  _ o 

+ dx2  + {y  ’ dx 2 _ 


1 + 


Therefore, 


y — n—  — 


dx 2 


d?y 

dtf 


x — m — 


/ df\dy 
\ dxV  dx 


d^y 

dx 2 


By  (5),  (8),  and  (9),  we  have 
r — ± 

From  (9)  and  (8)  we  have 


(i  - %r 


&y_ 

dx 2 


m = x 


dx2!  dx 


n = y + 


(Fy 
dx 2 

i + it 

+ dx 2 


cPy 
dx 2 


(5) 

(6) 

(7) 

(8) 
(9) 


(10) 


(11) 


(12) 


120.  Second  Method. — Let  ds  denote  an  infinitely 
small  element  of  a curve  at  a point,  and  <f>  the  angle  which 
the  tangent  at  this  point  makes  with  the  axis  of  x.  Imagine 
two  normals  to  be  drawn  at  the  extremities  of  this  elemen- 
tary arc,  i.  e.,  at  two  consecutive  points  of  the  curve  ; these 


RADIUS  OF  CURVATURE. 


221 


normals  will  generally  meet  at  a finite  distance.  Let  r be 
the  distance  from  the  curve  to  the  point  of  intersection  of 
these  consecutive  normals.  Then  the  angle  included  be- 
tween these  consecutive  normals  is  equal  to  the  correspond- 
ing angle  of  contingence  (Art.  116),  i.  e.,  equal  to  d<f>.  Since 
dcp  is  the  arc  between  the  two  normals  at  the  unit’s  distance 
of  the  point  of  intersection,  we  have 

( 7<? 

ds  — r d<p,  or  r — (1) 

Now  this  value  of  r evidently  represents  the  radius  of  the 
circle,  which  has  the  same  curvature  as  that  of  the  given 
curve  at  the  given  point,  and  hence  is  the  radius  of  curva- 
ture for  the  given  point,  while  the  centre  of  curvature  may 
be  defined  as  the  point  of  intersection  of  two  consecutive 
normals. 


To  find  the  value  of  r,  we  have  (Art.  56a), 

— tan-1  — • 


tan  <p  = ^-  ; 
dx 


<P 


dx  ’ 


and  hence  d(p  = 


dfy 

dx 


i + ar 

+ dx* 


also,  ds  — dx?  + dy2. 


Substituting  in  (1),  we  have 


r — 


(i  + 

V + dx2) 


dfy 
dx 2 


(2) 


which  is  the  same  as  (10)  of  Art.  119. 

/ dy2\* 

As  the  expression  ^1  + has  always  two  values,  the 

one  positive  and  the  other  negative,  while  the  curve  can 
generally  have  only  one  definite  circle  of  curvature  at  any 
point,  it  will  be  necessary  to  agree  upon  which  sign  is  to  be 


222 


RADIUS  OF  CURVATURE. 


taken.  We  shall  adopt  the  positive  sign,  and  regard  r as 
positive  when  the  second  derivative  is  positive,  i.  e.,  when 
the  curve  is  convex  downwards.  (Usage  is  not  uniform  on 
this  point.  See  Price’s  Calculus,  Vol.  I,  p.  435.  Todhun- 
teris  Calculus,  p.  339,  etc.) 

121.  To  Find  the  Radius  of  Curvature  in  Terms 
of  Polar  Co-ordinates. 

We  may  obtain  this  by  transforming  (2)  of  Art.  120  to 
polar  co-ordinates,  from  which  we  find 


where  N is  the  normal.  See  Art.  102,  Eq.  9.  [See  (2)  of 
Ex.  4,  Art.  90.] 

122.  At  a Point  where  the  Radius  of  Curvature 
is  a Maximum  or  a Minimum,  the  Circle  of  Curva- 
ture has  Contact  of  the  Third  Order  with  the  Curve. 

Since  r is  to  be  a maximum  or  a minimum,  we  must 

have  — 0. 
ax 

Differentiating  (2)  of  Art.  120  with  respect  to  x,  we  have 


... 

*•  da* 


odyttW 
° dx\dxV 


1 + 


dtf 

da* 


(1) 


DIFFERENT  ORDERS  OF  CONTACT. 


223 


Differentiating  (8)  of  Art.  119,  we  have 

cly  /cPy^ 

d3y  _ dx\dx2)  .0. 

do*  ~ ~ df'  ^ 

1 + dx2 

Hence  the  third  derivative  at  a point  of  maximum  or 
minimum  curvature  is  the  same  as  it  is  in  the  circle  of 
curvature,  and  therefore  the  contact  at  this  point  is  of  the 
third  order  (Art.  117). 

Cor. — The  contact  of  the  osculating  circles  at  the 
vertices  of  the  conic  sections  is  closer  than  at  other 
-points. 

123.  Contact  of  Different  Orders. — Let  y = f(x) 
and  y — <p  ( x ) represent  two  curves,  and  let  x , be  the  ab- 
scissa of  a point  of  their  intersection ; then  we  have 

f{xx)  = 0 (z,). 

Substituting  xx  ± h for  z,  in  both  equations,  and  sup- 
posing yx  and  y.2  the  corresponding  ordinates  of  the  two 
curves,  we  have 

* =/(*  ± >‘)  =/(*.)  +/'(*.)  (±i) 

+/"'W%|-  + etc.  (1) 

y,  =♦(*■  ± }l)  = ’Kzi)  + (±  0 + 4>"  (*■)  * ±f- 

+ <t'"  0.)  gfg  + etc.  (2) 

Subtracting  (2)  from  (1),  we  get,  for  the  difference  of 
their  ordinates,  corresponding  to  Zi  ± h, 

Vi-y,=  [f(xO-cp'(x  0]  (±h)  + [/"(*,)  - 0"(z,)] 

+ [/"'  (Xl)  - 6"'  (z,)]  + etc.  (3) 


224 


EXAMPLES. 


Now  if  these  curves  have  contact  of  the  first  order,  the 
first  term  of  (3)  reduces  to  zero  (Art.  117).  If  they  have 
contact  of  the  second  order,  the  first  two  terms  reduce  to 
zero.  If  they  have  contact  of  the  third  order,  the  first  three 
terms  reduce  to  zero,  and  so  on.  Hence,  when  the  order  of 
contact  is  odd,  the  first  term  of  (3)  that  does  not  reduce  to 
zero  must  contain  an  even  power  of  ± h,  and  y,  — y3  does 
not  change  sign  with  h,  and  therefore  the  curves  do  not 
intersect,  the  one  lying  entirely  above  the  other  ; but  when 
the  order  of  contact  is  even,  the  first  term  of  (3)  that  does 
not  reduce  to  zero  must  contain  an  odd  power  of  ± h,  and 
2/i  — y-i  changes  sign  with  h,  and  therefore  the  curves  inter- 
sect, the  one  lying  alternately  above  and  below  the  other. 

Cor.  1. — At  a point  of  inflexion  of  a curve,  the  second 
derivative  equals  0 ; also,  the  second  derivative  of  any  point 
of  a right  line  equals  0.  Hence,  at  a point  of  inflexion , 
a rectilinear  tangent  to  a curve  has  contact  of  the 
second  order,  and  therefore  intersects  the  curve. 

Cor.  2. — Since  the  circle  of  curvature  has  a contact  ol 
the  second  order  with  a curve,  it  follows  that  the  circle  of 
curvature,  in  general,  cuts  the  curve  as  well  as 
touches  it. 

Cor.  3. — At  the  points  of  maximum  and  minimum  curva- 
ture, as  for  example  at  any  of  the  four  vertices  of  an  ellipse, 
the  osculating  circle  does  not  cut  the  curve  at  its  point  of 
contact. 


EXAMPLES. 


1.  Find  the  radius  of  curvature  of  an  ellipse, 


a2  + 


dy  _ Vx  > 

dx  — a2y  ’ 


dif  _ edy2  4- 
dx2  ~ cdy2 


Here 


EXAMPLES. 


225 


cdy 2 


l2  (a?y2  -f-  tike2}  b* 

ay  ~ ~ a2 f 


(Art.  120),  r = 


<Py 

dx2 


(ay  + lAx2)%  . 

= (neglecting  the  sign), 


At  the  extremity  of  the  major  axis, 
x = a,  y = 0, 

At  the  extremity  of  the  minor  axis, 
x — 0,  y = l, 


V 

r = — • 
a 


a * 

r = b' 


2.  Find  the  radius  of  curvature  of  the  common  parabola. 


Here 


dy  = p 
dx  y’ 


y2  — 2px. 

cPy 
dx2  ~ 


y 


rS  ’ 


(y2 -\- p2)^  (normal)8 
- ^ “ 


f 

At  the  vertex,  y — 0 ; r = p. 

3.  Find  the  radius  of  curvature  of  the  cycloid 

x — r vers-1  ^ — V 2 ry  — y2. 

Here  -- 9 • l + ^-t 

dy  ~ *•  + da*  - y : 


cPv 

dtf 


y 


.2  » 


r = 2V2ry, 


which  equals  twice  the  normal  (Art.  101,  Ex.  51. 


226 


EVOLUTE  S AND  INVOLUTES. 


4.  Find  the  radius  of  curvature  of  the  parabola  whose 
latus-rectum  is  9,  at  x — 3,  and  the  co-ordinates  of  the 
centre  of  curvature.  r — 16.04;  m — 13-|-,  n = — 6.91. 

5.  Find  the  radius  of  curvature  of  the  ellipse  whose  axes 

are  8 and  4,  at  x — 2,  and  the  co-ordinates  of  the  centre  of 
curvature.  r = 5.86;  m = .38,  n = — 3.9. 

6.  Find  the  radius  of  curvature  of  the  logarithmic  spiral 

r — ae. 


dr 

dd 


— a 0 log  a ; 


d2r 

W2 


/70 


a 6 log2  a ; 


R = 


(r2  + r2  log2  a)  - / 2 , 2 1 2 \4  nr 

— 9 = (r2  + r2  log2  a)^  = K 
r2  + 2 r2  log2  a — r2  log2  a 

(See  Ex.  2,  Art.  102.) 

7.  Find  the  radius  of  curvature  of  the  spiral  of  Archi- 


medes, r — aQ. 


R = 


(a2  -\-  r2) 2 


2a2  + r 2 

8.  Find  the  radius  of  curvature  of  the  hyperbolic  spiral, 

r (a2  + r2)^ 


rd  = a . 


R = 


124.  Evolutes  and  Involutes. — The  curve  which  is 
the  locus  of  the  centres  of  all  the  osculating  circles  of  a 
given  curve,  is  called  the  evolute  of  that  curve  } the  latter 
curve  is  called  the  involute  of  the  former. 

Let  P„  P2,  P3,  etc.,  represent  a series  of  M 
consecutive  points  on  the  curve  MU’,  and 
C„  Cs,  C3,  etc.,  the  corresponding  centres 
of  curvature  ; then  the  curve  C„  C2,  C3, 
etc.,  is  the  evolute  of  MN,  and  MN  is  the 
involute  of  C„  C2,  C3,  etc.  Also,  since  the 
lines  CiP„  C2P2,  etc.,  are  normals  to  the 
involute  at  the  consecutive  points,  the 
points  Cl5  C„  C3,  etc.,  may  be  regarded  as 


Fig.  38. 


EQUATION  OF  THE  E VOLUTE. 


227 


consecutive  points  of  the  evolute  ; and  since  each  of  the 
normals  PiC„  P2C2,  etc.,  passes  through  two  consecutive 
points  on  the  evolute,  they  are  tangents  to  it. 

Let  r1}  r2,  r3,  etc.,  denote  the  lengths  of  the  radii  of 
curvature  at  P„  P2,  P3,  etc.,  and  we  have, 

r2-rx  = P2C2  - P.C,  = P2C2  - PaC,  = C,C2 ; 

also  r3  - r2  = P3C3  - P2C2  = P3C3  - P302  = C2C3 ; 

and  r4  — r3  = C3C4,  and  so  on  to  r„ ; 

hence  by  addition  we  have, 

rn  — r,  = CtC2  -f-  C2C3  + • • • ■ a_,Cn. 

This  result  holds  when  the  number  n is  increased  indefi- 
nitely, and  we  infer  that  the  length  of  any  arc  of  the 
evolute  is  equal,  in  general,  to  the  difference  between 
the  radii  of  curvature  at  its  extremities. 

It  is  evident  that  the  involute  may  be  generated  from  its 
evolute  by  winding  a string  round  the  evolute,  holding  it 
tight,  and  then  unwinding  it,  each  point  in  the  string 
will  describe  a different  involute.  It  is  from  this  property 
that  the  names  evolute  and  involute  are  given.  While  a 
curve  can  only  have  one  evolute,  it  can  have  an  infinite 
number  of  involutes. 

The  involutes  described  by  two  different  points  in  the 
moving  string,  are  said  to  be  parallel ; each  curve  being  got 
from  the  other  by  cutting  off  a constant  length  on  it? 
normal,  measured  from  the  involute.  (Williamson’s  Dif- 
ferential Calculus,  p.  295.) 

125.  To  find  the  Equation  of  the  Evolute  of  a 
Given  Curve. — The  co-ordinates  of  the  centre  of  curva- 
ture are  the  co-ordinates  of  the  evolute  (Art.  124).  Hence, 
if  we  combine  (11)  and  (12)  of  Art.  119  with  the  equation 
of  the  curve,  and  eliminate  x and  y,  there  will  result  an 
equation  expressing  a relation  between  m and  n,  the  co-or- 


228 


EXAMPLES. 


dinates  of  the  required  e volute,  which  is  therefore  the 
required  equation  ; the  method  can  be  best  illustrated  by 
examples. 

The  eliminations  are  often  quite  difficult;  the  following 
are  comparatively  simple  examples. 

EXAMPLES. 

1.  Find  the  equation  of  the  evolute  of  the  parabola, 


y = — p7inS. 

And  these  values  of  x and  y in  (1)  give, 

= I P(m  —p)\ 


n2  = (2) 

which  is  the  equation  required,  and  is  called  th e semi-cubical 
parabola.  Tracing  the  curve,  we  find  its  form  as  given  in 
Fig.  39,  where  AO  = p. 

If  we  transfer  the  origin  from  O to  A,  (2)  becomes 


EXAMPLES. 


229 


2.  Find  the  length  of  the  evolute  AQ',  Fig.  39,  in  terms 
of  the  co-ordinates  of  its  extremities. 

Let  ON  = x,  NQ  = y ; ON'  - m,  N'Q'  = n. 

Then  by  Art.  123,  Ex.  2,  we  have, 

r _ 

p 2 

Therefore,  by  Art.  124,  we  haye. 

Length  of  AQ'  = Q Q — AO  = + _p 

p 4 

— (ni  -j-  pi)%  — p.  (Since  y 2 = p^vX,  by  Ex.  1.) 


3.  Find  the  equation  of  the  evolute  of  the  cycloid, 


x = r vers' 


_1  - — \/  2 ry  — y2. 


(D 


dy  A /%ry  — y2  d2y  r 

dx  y ’ dx2  y2 

m = x - f 2 \/'2ry  — y 2 and  n = — y ; 

x — m — 2 a/  — 2 rn  — n 2 and  y = — n ; 

which,  in  the  equation  of  the  cycloid,  gives 

m = r vers- 1 ( — + *J—2rn—n2,  (2) 

V 

which  is  the  equation  of  a cy- 
cloid equal  to  the  given  cycloid ; 
the  origin  being  at  the  highest 
point.  This  will  appear  by  re- 
ferring the  given  cycloid  OO'B 
to  the  parallel  axes  O'X'  and 
O'Y'. 


Here 


or 


230 


PROPERTIES  OF  EYOLUTES. 


Let  (m,  n)  be  any  point  P in  the  cycloid ; then  we  have 
n — — MP  = — HD, 
m O'M  = PP  — HB  — CB  + DP 
___  arc  PH  + DP 

= r vers-i  — J + x DO 

= r vers-1  ^ — — n (2r  _p  nf 

.-.  m — r vers-1  -)  + V — 2 rn  — n~ ; (4) 


which  is  the  same  as  equation  (2).  Hence  we  see  that  the 
equation  of  the  evolute,  OA  (2),  is  the  same  as  that  of  the 
cycloid,  O'B  (4).  That  is,  the  evolute  of  a cycloid  is  an 
equal  cycloid* 

4.  Find  the  equation  of  the  evolute  of  the  equilateral 
hyperbola  2 xy  = a>.  Am_  (m  + 


[Here  find  m + n — * ' t - 1 , 

L a2 

= etc.] 


m — n 


126.  A Normal  to  an  Involute  is  Tangent  to  th6 
Evolute. — This  was  shown  geometrically  in  Art.  124.  It 
may  also  be  shown  as  follows  : 

Let  (x,  y)  be  any  point  Q of  the  involute  (Fig.  40),  from 
which  the  normal  QQ'  is  drawn,  and  let  ( m , n)  be  the  point 
Q'  on  the  evolute  through  which  the  normal  passes. 

The  equation  of  QQ'  is 

y — n — — ^ (x  — m) ; (1) 


or 


x — m + (v  — n)  = 0. 
dx 


(2) 


How  when  we  pass  from  a point  Q,  to  a consecutive 
point  on  the  involute,  Q'  also  will  change  to  a consecutive 


* This  property  was  first  discovered  by  Huygens. 


ENVELOPES  OF  CURVES. 


231 


point  of  the  evolute,  therefore  we  differentiate  (2)  with 
respect  to  x,  regarding  x,  y,  m,  n,  as  variables,  and  get 


1 


dm 

dx 


+ S'  (!/-*)  + 


dtf 
dx 2 


dy  dn 
dx  dx 


(3) 


But,  since  (m,  n)  is  the  centre  of  curvature  corresponding 
to  Q,  we  have,  by  (8)  of  Art.  119, 


~ n)  % 


+ 1 + 


dtf  _ 


dx • 


= 0 


which  in  (3)  gives 


dm  dy  dn  _ dx dn 

dx  dx  dx  ’ °r  dy  dm  ’ 


and  this  in  (1)  gives 

dn  , , 

y-”  = 


(4) 


therefore  (1)  or  (4),  which  is  the  equation  of  a normal  to 
the  involute  at  Q ( x , y),  is  also  the  equation  of  a tangent  to 
the  evolute  at  Q'  ( m , n ). 

127.  Envelopes  of  Curves. — Let  us  suppose  that  in 
the  equation  of  any  plane  curve  of  the  form 

/ (*,  V,  a)  = 0,  (1) 

we  assign  to  the  arbitrary  constant  a,  a series  of  different 
values,  then  for  each  value  of  a we  get  a distinct  curve, 
different  from  any  of  the  others  in  form  and  position,  and 
(1)  may  be  regarded  as  representing  an  indefinite  number 
of  curves,  each  of  which  is  determined  when  the  correspond- 
ing value  of  a is  known,  and  varies  as  a varies. 

The  quantity  a is  called  a variable  parameter,  the  name 
being  applied  to  a quantity  which  is  constant  for  any  one 
curve  of  a series,  but  varies  in  changing  from  one  curve  to 
another,  and  the  equation  f{x,  y,  a)  = 0,  is  said  to  repre- 
sent a family  of  curves. 

If  we  suppose  a to  change  continuously,  i e..  by 


232 


EQUATION  OF  THE  ENVELOPE. 


infinitesimal  increments,  the  curves  of  the  series  represented 
by  (1)  will  differ  in  position  by  infinitesimal  amounts  ; and 
any  two  adjacent  curves  of  the  series  will,  in  general,  inter- 
sect; the  intersections  of  these  curves  are  points  in  the 
envelope.  Hence  an  envelope  may  be  defined,  as  the  locus 
of  the  intersection  of  consecutive  curves  of  a series. 

It  can  be  easily  seen  that  the  envelope  is  tangent  to  each 
of  the  intersecting  curves  of  the  series;  for,  if  we  consider 
four  consecutive  curves,  and  suppose  P,  to  be  the  point  of 
intersection  of  the  first  and  second,  P2  that  of  the  second 
and  third,  and  P3  that  of  the  third  and  fourth,  the  line 
Pj  P2  joins  two  infinitely  near  points  on  the  envelope  and 
on  the  second  of  the  four  curves,  and  hence  is  a tangent 
both  to  the  envelope  and  the  second  curve  ; in  the  same 
way  it  may  be  shown  that  the  line  P2P3  is  a tangent  to  the 
envelope  and  the  third  consecutive  curve,  and  so  on. 


128.  To  Find  the  Equation  of  the  Envelope  of  a 
given  Series  of  Curves. 

Let  f(x,  y,  a)  = 0,  (1) 

f(x,y,a  + da)  = 0,  (2) 


be  the  equations  of  two  consecutive  curves  of  the  series ; 
then  the  co-ordinates  of  the  points  of  intersection  of  (1)  and 
(2)  will  satisfy  both  (1)  and  (2),  and  therefore  also  will 
satisfy  the  equation 

/(x,y,<1>~i('T’y’°+fe)'  = 0 <*»-•. Art-  30>’ 


or 


df(x,  y,  a)  _ 

da  ~ ’ 


(3) 


and  therefore  the  points  of  intersection  of  two  infinitely 
near  curves  of  the  series  satisfy  each  of  the  equations  (1) 
and  (3).  Hence,  to  find  the  equation  of  the  envelope,  we 
eliminate  a between  (1)  and  (3),  i.  e.,  we  eliminate  the  varia- 
ble parameter  between  the  equation  of  the  locus  and  its  first 
differential  equation. 


EXAMPLES. 


233 


EXAMPLES. 


1.  Find  the  envelope  of  y = ax  q , when  a varies. 

a 

Differentiating  with  respect  to  a,  x,  and  y,  being  constant, 
we  have 


,\  y — ± [V mx  -f  V mx\  or  y 2 = 4 mx, 
which  is  the  equation  of  a parabola. 


2.  A right  line  of  given  length 
slides  down  between  two  rectangular 
axes ; to  find  the  envelope  of  the  line 
in  all  positions. 

Let  c be  the  length  of  the  line,  a 
and  b the  intercepts  OA  and  OB ; 
then  the  equation  of  the  line  is 


? , y _ 

a'b~~ 


L 


(1) 


in  which  the  variable  parameters  a and  b are  connected  by 
the  equation 

a2  -f  b2  = c2.  (2) 

Differentiating  (1)  and  (2),  regarding  a and  b as  varia- 
ble, we  have 


--.da-  | db  = 0,  or  - -da 
a2  o2  a2 


o db. 


ada  + bdb  — 0,  or  — ada  — bdb. 


Dividing  (3)  by  (4),  we  have 

x y x y 

x y a __  b _ a b 1 

a3  _ i®  ’ °r  "a2  ~ ~ a*  + ¥ = ? 5 

a = (zc2)^  and  b — ( yc 2)L 


(3) 

W 


234 


EXAMPLES. 


which  in  (2)  gives 

(ze2)s  + {yc2)3  = c2; 

2 2 2 
+ y*  = c3, 

which  is  the  equation  required. 
The  form  of  the  locus  is  given  in 
Fig.  42,  and  is  called  a hypo-cycloid , 
which  is  a curve  generated  by  a 
point  in  the  circumference  of  a 
circle  as  it  rolls  on  the  concave 
arc  of  a fixed  circle. 


3.  Find  the  envelope  of  a series  of  ellipses  whose  axes  are 
coincident  in  direction,  their  product  being  constant. 


Here 

X 2 

e* 

+ 

11 

1. 

(1) 

Let 

a 

■b  = 

c ; 

(2) 

• 

• • 

x2  , 
o da  + 
a 3 

Pl 

o, 

or 

X2 

as 

da  -- 

-Pk 

(3) 

da 

db 

o. 

da 

db 

a 

+ T 

— 

or 

a 

b * 

(4) 

Dividing  (3)  by  (4),  we  have 

y2 

-5  = 75  = 4,  by  substituting  in  (1). 
a2  b2 

a = ± xV2,  and  b — ± yV%’, 

from  (2),  xy  = ± |> 

which  is  the  equation  of  an  hyperbola  referred  to  its  asymp- 
totes as  axes. 

This  example  may  also  be  solved  as  follow's : Eliminating 
l from  (1)  and  (2),  we  have 


Fig.  42. 


EXAMPLES. 


235 


t ^ 

a2+  c2 


I/2  = l, 


(5) 


in  which  we  have  only  the  variable  parameter  a. 

. _^2  ,^_0.  • a2--- 

c3  “ u’  ‘ y’ 


(6) 


winch  in  (5)  gives 


Xy  + XJL=  l;  ...**  = * 


4.  Find  the  envelope  of  the  right  lines  whose  general 
equation  is 

y =z  mx  + (ahn2  + b2)*,  (1) 

where  m is  the  variable  parameter. 


We  find  m = — - — , 

a 's/  a2  — x 2 

which  in  (1)  gives  — + ^ = 1 for  the  required  envelope. 

Hence  the  envelope  of  (1)  is  an  ellipse,  as  we  might  have 
inferred,  since  (1)  is  a tangent  to  an  ellipse.  (See  Anal. 
Geom.,  Art.  74.) 


EXAMPLES. 


1.  Find  the  radius  of  curvature  of  the  logarithmic  curve 


Z = log  y. 


_ ( m 2 + y2)? 

my 


2.  Find  the  radius  of  curvature  of  the  cubical  parabola 


j2  --  a2x. 


I (V  + a <)» 

6aAy 


3.  Find  the  radius  of  curvature  of  the  curve 


y — x3  — x2  -4-  1 


236 


EXAMPLES. 


where  it  cuts  the  axis  of  y,  and  also  at  the  point  of  mini 
mum  ordinate. 

At  the  first  point,  r = — ; at  the  second,  r = 

4.  Find  the  radius  of  curvature  of  the  curve 

yz  = 6x2  + or?. 

_ [y4  + (4g  + 

? ~ -8  x2y 

5.  F’.nd  the  radius  of  curvature  of  the  rectangular  hyper- 
bola zy  = m*.  (a;2  + ^2)1 

T 2w2 

6.  Find  the  radius  of  curvature  of  the  Lemniscate  oi 

Bernouilli  r2  = a?  cos  26.  _ a2 

K ~~  3r‘ 

7.  Find  the  equation  of  the  evolute  of  the  ellipse 

a2y2  + b2x2  = a?b2. 

(am)%  + (bn)i  = (a2  — J2)*. 

8.  Find  the  equation  of  the  evolute  of  the  hyperbola 

a?y 2 — b2x2  = — a2b2. 

(am)i  — (bn)^  = ( a 2 + b2)*. 

9.  Prove  that,  in  Fig.  39,  OM  = 40A  = 4p,  and  MP 

= 2pV2. 

10.  Find  the  length  of  the  evolute  AP'  in  Fig.  39. 

A ns.  (3?  — 1 )p. 

11.  Find  the  length  of  the  evolute  of  the  ellipse.  (See 

Art.  123,  Ex.  1,  and  Art.  124.)  ^ ^ a?  — b 3 

ab 

12.  Find  the  length  of  the  cycloidal  arc  OO'B,  Fig.  40. 

A ns.  8 r. 

13.  Find  the  envelope  of  the  series  of  parabolas  whose 
equation  is  y2  = on  ( x — on),  on  being  the  variable  parameter 


EXAMPLES. 


237 


14.  Find  die  envelope  of  the  series  of  parabolas  expressed 


by  the  equation  y — ax —~x2,  where  a is  the  variable 

parameter. 

The  result  is  a parabola  whose  equation  is 


a8  = (|  — y\ 


This  is  the  equation  of  the  curve  touched  by  the  parabolas  de- 
scribed by  projectiles  discharged  from  a given  point  with  a constant 
velocity,  but  at  different  inclinations  to  the  horizon.  The  problem 
was  the  first  of  the  kind  proposed,  and  was  solved  by  John  Bernouilli, 
but  not  by  any  general  method. 

15.  Find  the  envelope  of  the  hypothenuse  of  a right- 

angled  triangle  of  constant  area  c.  c 

Xy  ~ 2 

16.  One  angle  of  a triangle  is  fixed  in  position,  find  the 
envelope  of  the  opposite  side  when  the  area  is  constant  — c. 

c 

Xy  ~ 2* 

17.  Find  the  envelope  of  x cos  a + y sin  a = p,  in 

which  a is  the  variable  parameter.  x2  + y2  — p2. 


18.  Find  the  envelope  of  the  consecutive  normals  to  the 
parabola  y2  — 2px. 

g 

Ans.  y 2 = (x  — p)3,  which  is  the  same  as  was  found 

fiJi  P 

for  the  evolute  in  Ex.  1,  Art.  125,  as  it  clearly  should  be. 
(See  Art.  124.) 


19.  Find  the  envelope  of  the  consecutive  normals  to  the 
ellipse  a2y2  + i2x2  = a2l2. 

Ans.  ( ax )*  + (t>y)f  — («2  — b2)2,  which  is  the  same  as 
was  found  in  (7)  for  the  evolute  of  the  ellipse. 


PART  II. 

INTEGRAL  CALCULUS. 


CHAPTER  I. 


ELEMENTARY  FORMS  OF  INTEGRATION. 


129.  Definitions. — The  Integral  Calculus  is  the  inverse 
of  the  Differential  Calculus,  its  object  being  to  find  the 
relations  between  finite  values  of  variables  from  given 
relations  between  the  infinitesimal  elements  of  those  vari- 
ables ; or,  it  may  be  defined  as  the  process  of  finding  the 
function  from  which  any  given  differential  may  have  been 
obtained.  The  function  which  being  differentiated  pro- 
duces the  given  differential,  is  called  the  integral  of  the 
differential.  The  process  by  which  we  obtain  the  integral 
function  from  its  differential  is  called  integration. 

The  primary  problem  of  the  Integral  Calculus  is  to  effect 
the  summation  of  a certain  infinite  series  of  infinitesimals, 
and  hence  the  letter  S was  placed  before  the  differential  to 
show  that  its  sum  was  to  be  taken.  This  was  elongated 

into  the  symbol  f (a  long  S ),  which  is  the  sign  of  integra- 


tion, and  when  placed  before  a differential,  denotes  that  its 


integral  is  to  be  taken. 


3 x2dx,  which  is  read,  “ the 


integral  of  3 x*dx,”  denotes  that  the  integral  of  d'ddx  is  to 
be  taken.  The  signs  of  integration  and  differentiation  are 


ELEMENTARY  RULES  FOR  INTEGRATION. 


239 


inverse  operations,  and  when  placed  before  a quantity, 
neutralize  each  other.  Thus, 


and 


— ax, 


axdx  = axdx. 


130.  Elementary  Rules  for  Integration. — In  the  ele- 
mentary forms  of  integration,  the  rules  and  methods  are 
obtained  by  reversing  the  corresponding  rules  for  differ- 
entiation. When  a differential  is  given  for  integration,  if 
we  cannot  see  by  inspection  what  function,  being  differ- 
entiated, produces  it,  or  if  it  cannot  be  integrated  by  known 
rides,  we  proceed  to  transform  the  differential  into  an 
equivalent  expression  of  known  form,  whose  integral  we 
can  see  by  inspection,  or  can  obtain  by  known  rides.  In 
every  case,  a sufficient  reason  that  one  function  is  the 
integral  of  another  is  that  the  former,  being  differentiated, 
gives  the  latter.* 


(1.)  Since 

d (v  + y — z)  = dv  + dy  — dz\  (Art.  14.) 

J* ( dv  + dy  — dz)  = j' d (v  + y — z)  — v + y — z 

= I* dv  + j‘dy  — J dz. 


Hence,  the  integral  of  the  algebraic  sum  of  any 
number  of  differentials  is  equal  to  the  algebraic  sum 
of  their  integrals. 

(2.)  Since 

d (ax  ± b)  = adx ; (Art.  15.) 


* While  there  is  no  quantity  whose  differential  cannot  he  found,  there  is  a large 
class  of  differentials  whose  integrals  cannot  be  obtained  ; either  because  there  is  no 
quantity  which,  being  differentiated,  will  give  them,  or  because  the  method  for 
their  integration  has  not  yet  been  found. 


MO 


ELEMENTARY  RULES  FOR  INTEGRATION 


adx  = / d ( ax  -f  b)  = ax  ±b 


Hence.,  a constant  factor  can  be  moved  from  onu 
side  of  the  integral  sign  to  the  other  without  affect- 
ing the  value  of  the  integral.  Also,  since  constant 
terms,  connected  by  the  sign  ±,  disappear  in  differentia- 
tion, therefore  in  returning  from  the  differential  to  the 
integral,  an  arbitrary  constant,  as  C,  must  be  added,  whose 
value  must  be  determined  afterwards  by  the  data  of  the 
problem,  as  will  be  explained  hereafter. 


d \ L/»]n  = « [/(z)]"-1  df(x) ; (Arts.  15  and  19.) 

••  fa  [/(a;)]*-1  df(x)  = f d | [/(x)]n  = | [/(z)]n  + c. 


Hence,  ivhenever  a differential  is  the  product  of 
three  factors,  viz,  a constant  factor , a variable  factor 
with  any  exponent  except  — 1,  and  a,  differential 
factor  which  is  the  differential  of  the  variable  factor 
without  its  exponent,  its  integral  is  the  product  of 
the  constant  factor  by  the  variable  factor  with  its 
exponent  increased  by  1,  divided  by  the  new  ex- 
ponent.* 

It  will  be  seen  that  the  rule  fails  when  n = — 1,  since 
if  we  divide  by  1 — 1 = 0,  the  result  will  be  co . 


(f.)  Since  d ( a log  x)  = — - ; (Art.  20,  Cor.) 


* The  arbitrary  constant  is  not  mentioned  since  its  addition  is  always  under- 
stood, and  in  the  following  integrals  it  will  b<»  omitted,  as  it  can  always  be  supplied 
when  necessary. 


(3.)  Since 


adx 


d (a  log  x)  — a log  x. 


EXAMPLES. 


241 


Hence,  whenever  a differential  is  a fraction  whose 
numerator  is  the  product  of  a constant  by  the  differ- 
ential of  the  denominator , its  integral  is  the  product 
of  the  constant  by  the  JYaperian  logarithm  of  the 
denominator. 

EXAMPLES. 

1.  To  integrate  dy  = ax5dx. 

y = J* axhlx  — J a • x?  • dx  = [by  (5)]. 


2.  To  integrate  dy  = (a  + 5x?)ix2dx. 

The  differential  of  the  quantity  within  the  parenthesis 
being  15 x2dx,  we  write 

y — f is  («  + Sr5)4 15 x2dx  — (a  +7^--  [by  (5)], 

This  example  might  also  be  integrated  by  expanding  the  quantity 
within  the  parenthesis,  and  integrating  each  term  separately  by  (1), 
but  the  process  would  be  more  lengthy  than  the  one  employed. 


3.  To  integrate 

dy  — a (ax2  + bx?)?  2 xdx  + 3 bx2  (ax2  + bx?)?  dx. 


y — J ^ \a  (ax2  -f  bx3)?  2x dx  + 3 bx2  (ax2  + bx3)?  dx} 

= J * (ax2  + bx3)?  (2  ax  + 3 bx2)  dx  — §-  (ax?  + bx3)?  [by  (5)]. 


4. 


To  integrate  dy  = 


adx 
a + bx 


Since  the  numerator  must  be  bdx  to  be  the  differential  of 
the  denominator,  we  must  multiply  it  by  b,  taking  care  to 
divide  by  b also  ; hence, 


V = 


/ 


adx 
a + bx 


a 

~b 


f 


bdx 

a + bx 


= l log  (a  + bx).  [by  (If)}. 


11 


242 


DIFFERENTIAL  FORMS. 


131.  Fundamental  Forms. — On  referring  to  the  forms 
of  differentials  established  in  Chap.  II,  we  may  write  down 
at  once  the  following  integrals  from  inspection,  the  arbitrary 
constant  being  always  understood. 


1. 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 


10. 


11. 

12. 

13. 


axn 


n + 1 


y = J*  axndx 

f'adx 

y J %"■ 

/adx 

IT 

y = J n ax  log  a dx 
y = J cNdx 

y — f cos  a;  dx 
y = J — sin  x dx 
y = f sec2  a:  dx 
y — ) — cosec2  x dx 
y = ^sec  x tan  x dx 
y = f — cosec  x cotan  x dx  — cosec  x. 
y — f sin  x dx  - vers  x. 

y — f — cos  x dx  = covers  x. 


( n — 1)  zn_1 
= a log  x.* 

= ax. 

= c®. 
i sin  x. 

— COS  X. 

= tan  x 
--  cot  X. 

: sec  X. 


* Since  the  constant  c to  be  added  is  arbitrary,  log  c is  arbitrary,  and  we  may 
K-rite  the  integral  in  the  form 


a log  x + log  c = log  caf. 


INTEGRATION  BY  TRANSFORMATION. 


243 


14. 

15. 

16. 


17. 


18. 


19. 


20. 

21. 


y 


=/ 

=/- 

=/i 


dx 

Vi  — ^ 

dx 


VI 

dx 


y = f 

!)=/ 


+ a:2 
dx 

1 + z2 
dx 


= In. 

= /■ 


a;  V z2  — 1 

dx 

xV  x3  — 
dx 
V&c  — a:2 
dx 


\/’2x  — a:2 


= sin-1  a;. 

— cos-1  a;. 

— - tan-1  x. 

— cot-1  a;. 

— sec-1  x. 

— cosec-1  x. 

— vers-1  x. 

— covers-1  x. 


These  integrals  are  caller  the  fundamental  or  elementary 
forms,  to  which  all  other  forms,  that  admit  of  integration 
in  a finite  number  of  terms,  can  be  ultimately  reduced.  It 
is  in  this  algebraic  reduction  that  the  chief  difficulty  of  the 
Integral  Calculus  is  found  ; and  the  processes  of  the  whole 
subject  are  little  else  than  a succession  of  transformations 
and  artifices  by  which  this  reduction  may  be  effected.  The 
student  must  commit  these  fundamental  forms  to  memory; 
they  are  as  essential  in  integration  as  the  multiplication 
table  is  in  arithmetic. 


132.  Integration  of  other  Circular  and  Trigono- 
metric Functions  by  Transformation  into  the  Fun- 
damental Forms. 

(It 

1.  To  integrate  dy  — 

V a3  — IP-x3 

We  see  that  this  has  the  general  form  of  the  differential 


244 


INTEGRATION  BY  TRANSFORMATION. 


of  an  arc  in  terras  of  its  sine  (see  form  14  of  Art.  131) ; 
hence  we  transform  our  expression  into  this  form,  as  follows: 


y 


[*  dx  _ p 

~ J y/a2  — b2x2  ~ J 


dx 


1 - 


Wx2 


= / 


- dx 
a 


A 


b2x2 


a “ V a i 

To  make  this  quantity  the  differential  of  an  arc  in  terms 
of  its  sine,  the  numerator  must  be  the  differential  of  the 
square  root  of  the  second  term  in  the  denominator,  which 

is  ^ dx.  Therefore  we  need  to  multiply  the  numerator  by  b, 

which  can  be  done  by  multiplying  also  by  the  reciprocal  of 
b,  or  putting  the  reciprocal  of  b outside  the  sign  of  integra- 
tion. Hence, 


= / 


dx 


y/  a2  — b2x2 


= / 


- dx 
a 


= l f- 

/ b2x2  / 

V1-^  V 


dx 


brX- 

~ If 


1 . , bx 

= T sin-1  — • 

b a 


2.  To  integrate  dy  = 

y = f 


dx 


Here 


V a2  — Vx2 


y/  a2  — W-x2 

dx  = /- 


dx 


L lb 

“V1- * 


2 


AS 


— -dx 
a 


1 - 


Vx2 


T COS' 
0 


bx 


S.  y 


= /, 


dx 


a2  + b2x2 
dx 


1 bx 

—=  tan  1 — • 
ab  a 


*•  » = /“?: 


_ _1  ,_1  bx 

+ b2x2  ~ ab  a ' 


PROCESS  OF  INTEGRATION. 


245 


s.  y = f 

6 ■ » = / 

7.  y 

8-  y = f 


dx 


xV  fix1  — a 2 
dx 

xVb'X2  — a2 

/dx 

\/'Zabx  — b2x2 
dx 


1 bx 

= - sec-1  — • 

a a 

1 bx 

— - cosec-1  — • 

a a 

1 , bx 

= T vers-1  - • 
b a 


1 , bx 

— y covers-1 — 

V 2 abx  — b2x2  b a 


^ P.  i P&mxdx  / 

9.  v=  tan  x dx  = / — — 

* J J COS  X J 


Cd  cos  z 


cos  x 


= — log  cos  a;  = log  sec  x. 


P , J /’cos  x dx  . 
10.  y = J cot*  dx  = j -yjjy-  = log 

_ P dx  _ P dx 

y J sin  x J 2 sin  A 


sin  *. 


[x  cos  lx 


=p 

»■  »-/£  = / 


sec2 (lx)  dx 
tan  £* 
dx 


= log  tan  J*. 


e+«) 

= log  tan  (j  + |x)  [by  (11)]. 


. « P dx  P 

13.  v — = / 

* J sm  x cos  x v 

dx 


sec3  x dx  , 

= log  tan  x. 

tan  x 

(sin2  x + cos2  x)  dx 


sura-  cos2* 


H.  y = = P 

y -J  sm2  x cos2  x J 

(since  sin2*  + cos2*  = 1)  = J* (sec2*  4-  cosec2 x)  dx 


246 


EXAMPLES. 


16.  y — J* cot2  x dx  — j* (cosec2  x — 1 )dx 

— — cot  X — X. 

17.  y = /’cos2  x dx  — j* (|  + \ cos  2x)  dx  (by  Trig. 

= \x  + \ sin  2x. 

(See  Price’s  Calculus,  Vol.  II,  p.  68.) 

18.  y = J* sin2  x dx  — \x  — J sin  2x. 


Remark. — It  will  be  observed  that  in  every  case  we  reduce  the 
function  to  a known  form,  and  then  pass  to  the  integral  by  simple 
inspection  or  by  the  elementary  rules.  Whenever  there  is  any  doubt 
as  to  whether  the  integral  found  be  correct  or  not,  it  is  well  to  differen- 
tiate it,  and  see  if  it  gives  the  proposed  differential.  (See  Art.  130.) 


EXAMPLES. 


3.  Ry  = bxJ  dx. 

Here  y = f* bx^  dx  =.  f* b-xh-dx 

— | fcct  [by  (3)  of  Art.  130]. 
xdx 


2.  dy  = 


Here 


V a 2 -f  x 2 


/xdx  n , 

jHa*  + a?)-l2xdz 


= (a2  + a:2)-. 

3. 

dy  = 2x%  dx. 

y = 

4. 

dy  = 2 x*  dx. 

II 

oi)ai 

w|ot 

5. 

II 

1 

i 

©x|3» 

£ 

y — 5x~i. 

INTEGRATING  FACTOR.  247 

7.  dy  = — 5 mx~*dx. 

8,  dy  — (f  cix?  — 5a;*)  dx 

9-  d,j= (§-!)*• 

(2  ax  — x 2)  . 

-r  dx. 

(3  ax2  — x3)  s 

j*  — (3 ax2  — xs)~  s (2 ax  — x 2)  f/a; 

= f — \{2>ax2 — a:3)- a (Qnx  — 3a;2) 

— — - tj  (3a:c2  — a:3)i 

11.  dy  = (2«a:  — x 2)*  (a  — a;)  dx.  y ■=.  \ (2ax  — x2)^. 

133.  Integrating  Factor. — It  has  been  easy,  in  the 
examples  already  given,  to  find  the  factor  necessary  to  make 
the  differential  factor  the  differential  of  the  variable  factor 
[see  (3)  of  Art.  130]  ; but  sometimes  this  factor  is  not  easily 
found,  and  often  no  such  factor  exists.  There  is  a general 
method  of  ascertaining  whether  there  is  such  a factor  or 
not,  and  when  there  is,  of  finding  it,  which  will  be  given 
in  the  two  following  examples : 

12.  dy  = 3 (45a;2  — 2cxff  (45a;  — 3ea;2)  dx. 

Suppose  A to  be  the  constant  factor  required ; then  we 
have 

y — f -*j-  (45a:2  — 2cz?)'s  (12.45a;  — 9 Acx2)  dx. 

If  A be  the  required  factor,  we  must  have 

d (45a:2  — 2ca;3)  = (\2Abx  — 9Acx2)  dx, 
or  85a;  — 6cx 2 = 1245a:  — 9 Acx2, 

and  since  this  result  is  to  be  true  for  every  value  of  x,  the 
coefficients  of  the  like  powers  of  x must  be  equal  to  each 


10.  dy  = 
Here  y = 


y = — 

y = off  — 5a:*. 

12  1 
y = - — + 3 


348 


EXAMPLES. 


other,  from  the  principle  of  indeterminate  coefficients,  giv- 
ing us  two  conditions, 

12.46  = 86,  (1) 

and  — 9 Ac  = — 6c,  (2) 

or  A = §.  from  (1),  and  A = §,  from  (2)  ; 
hence  § is  the  factor  required,  and  we  have 

y — j*  \ (46x2  — 2cx3)i  (86x  — Gcx2)  dx 
= f (46x2  — 2cxs)K 

13.  dy  — (2 ax  — x2p  (5a  — x)  dx. 

Let  A be  the  required  factor,  then 
p\  3 

y — J -j  (2 ax  — x2)^  (5.4a  — .4x)  dx. 

d (2 ax  — x2)  = (5.1a  — Ax)  dx  ; 
or  2 a — 2x  = 5Aa  — *4x. 

2a  = 5.4a  ; (1) 

and  — 2 = — A ; (2) 

or  A — §,  from  (1),  and  A = 2,  from  (2). 

These  values  of  A being  incompatible  with  each  other, 
we  infer  that  the  differential  cannot  be  integrated  in  this 
form. 


14. 


dy  = (26  + 3ax2  — 5x®)  » (2ax  — 5x2)  dx. 
y = \ (26  + 3ax2  — 5x3)s. 

15.  dy  = (3 ax3  + 46x®)*  (2 ax  + 46x2)  r/x. 

y = \ (3 ax2  + 46x®)h 

16.  % = atfdx  q — _ 

* 2\/x  ^3 


aX3  . A 

y = — + 


EXAMPLES. 


249 


17.  dy  — 5 (5a;2  — 2x)s  xdx  — (5a:2  — 2a:)  ^ dx. 

Here  y — J (5a:2  — 2x)i  (5x  — 1)  dx. 

y = \ (5z2  — a®)*. 

134.  A Differential  may  often  be  brought  to  the 

form  required  in  (3)  of  Art.  130,  by  transposing  a variable 
factor  from  the  differential  factor  to  the  variable  factor, 
or  vice  versa. 

, „ , 2 adx 

18.  dy  — « 

x v 2 ax  — x 2 

Here  y =■  J* (2 ax  — x2)~^  2 ax~xdx 

— j — (2ax~1  — l)~v  (_  2 ax~2)  dx ; 

. . y = — 2 (2ax~1  — l)v  [by  (3)  of  Art.  130] 

2 a/  2 ax  — x 2 


19.  rfy  = 

20.  dy  — 

21.  Jy  — 


a:r?a; 


(2«a;  — x2Y 
axdx 

(2  bx  + x2)§ 
adx 

x V-ibx  + 4c2a;a 


y = 
y = 


X 


a a/2  ax  — a:2 
ax 


b V%bx  + x2 


y = - 


2 a a/3  bx  + ‘kfix2 
3 bx 


dy  = xf-iiT^'  (4)  of  Art.  130.] 


J*d  (a  + bx  + ex2) 


23.  dy  = 


a + bx  + cx2 
8x3dx 


a + 2xt 


= log  (a  + bx  -f-  cx2). 

y = log  (a  + 2s4). 


250 


EXAMPLES. 


24. 


25. 


5 7?(kc 

du  = s*TT 
y = tv  log  (3xi  + 7) 

, \xdx 

d*  = ?+T 


= log  (3a^  + 7)A. 
y = log  (a;2  + f)i 


26. 

27. 


dy  — 


Ixclx 


8 a — 3x2 
dy  = (b  — x2)3x$dx. 


y = log 


1 

(8a  — 3a;2) 


y — | b3X?  — %b2X%  + fabx1*  — ^a^. 


7 

3 


7 xn~ldx  . , tJi 

28.  dy  = a - ixii-  y = log  (a  + bx")bn. 

dx 

29.  dy  — 2 log3  a;  — • V — \ log4  x- 

30.  dy  = a2*  log  adx.  [See  (4)  of  Art.  131.] 

Here  y — jo1*  log  adx  = f \ a & log  ad  (2x)  = | a 31 


31.  dy  — 3ax " log  axdx. 

32.  dy  — ba^dx. 

33.  dy  = 5 exdx. 

34.  dy  = cos  3 xdx.  [See  (6)  of  Art.  131.] 

y = ^ sin  3a:. 

35.  dy  = cos  (.r2)  xdx.  y — 4 sin  ( x 2). 

36.  dy  = e6in  x cos  xdx.  y = esin  x. 

37.  dy  = — ecosx  sin  xdx.  y = ecos  *. 

38.  dy  — sin2  (2a:)  cos  (2a:)  dx. 

Here  y — f J sin2  (2a:)  cos  (2a:)  d (2a:)  = £ sin3  (2a;). 


y = laX'- 

ba3* 

y ~ 3 Toga* 
y = 5e*. 


39. 

40. 


EXAMPLES. 


251 


dy 
dy 

41.  dy 

42.  dy 

43.  dy 

44.  dy 

45.  dy 

46.  dy 

47.  dy 
Here 

48.  dy 

49.  dy 


= cos3  (2a;)  sin  (2a;)  dx. 
= sec2  (x)3  xzdx. 

= sec  (3a;)  tan  (3a;)  <Za;. 

= sin  (aa;)  dx. 


V - cos4  (2a;). 

V = i tan  (a;)3. 
y = 4 sec  (3a;). 


y = - vers  (aa;). 


tan  xdx. 
sin  0 sec2  0fZ0. 

9/7t 

--  [See  (14)  of  Art.  131.1 

Vl-4a;2  L V J 


y = log  sec  ar. 
y — sec  0. 


axdx 


Vl  — z4 

x)'dx 

V2  — 4a;3 

* = / 


y = sin-1  (2a-). 
V = | sin-1  (a;2) 


xhlx 


a/2  • Vl  — 2 a3 

a/2  • \xhrix 


= — - f 
= */: 


V2  • a/1  — 2a^ 
d V2X3 


Vl  — 2x3 
y = i sin-1  V 2aA 
dx 


V2  — 9a2 
— xdx 


y — i sin'1 


3a 

v/ 2 


a/2  — 

dx 


y = 


2 a/5 


cos 


(A) 


a/  aa; 


50.  dy  = — 


y — 2 cos  1 


a. 


252 


EXAMPLES. 


51.  dy 

52.  dy 

53.  dy 

54.  dy 
Here  y 

55.  dy 

56.  eft/ 

5?. 

58.  c??/ 
69.  dy 

60.  dy 

61.  dy 

62.  dy 


3dx 

4 + 9x2 

1 , , 3x 

V = i tan-1  - • 

x2dx 
1 + ofi 

y = \ tan-1  x3. 

3dx 

a/6x  — 9x2 

y = vers-1  3x. 

2 dx 

x a/3x2  — 5 

P 2 dx 

2 /*  VlrZx 

x \/5  V |x2  1 V5  -\/-§x  a/ — 

2 

: V5  SeC_1  ^ ^ 

2 dx 

y — \/ 2 cosec-1  (x  Vf). 

< 

Oi 

1 

cS 

— 2x~xdx 

2 

y = — — cosec-1  (x  V-1^-). 

Vl4x2  — 3 

3xldx 

V 4x3  — 9x6 

y = i vers  1 — • 

— 6a</x 
V 6 — 9x2 

y — a covers-1  3X3. 

8x~^dx 
5/  2x^  — 6x5 

y = 4 a/6  vers-1  (6x3). 

1 — sin  x , 

— ax. 

x + cos  X 

y = log  (x  + cos  x). 

xdx 

1 + X4' 

y — 4 tan-1  x2. 

x2 — 4x  + 3 
x3 — 6x2  + 9x  ^X‘ 

y = log  (x3  — 6x2  4 9x)i 

EXAMPLES. 


253 


3 dx 

63,  dy  = 2 + 5x2' 


tan_1 27  ^ 


64.  rfy  = <*^fe.  v = i(a+Vif. 

J *Jx 

65.  ,j-=2xV*+^ 
V«2  + a;2 

66.  dy  — tan3  a:  sec3  xdx.  y — \ tan3  a:. 


67.  dy  = 


68.  dy  = 


(1  — a:3)3  rfx 


(a;  — 2)  dx 


* Vx 

, Mx 
69.  dy  - ^qrx' 


y = loga-^  + J* 

j,  = 2v^  + ~ 
y = ^x3  — * + tan_1x. 


70.  dy  = 


dx 


1 + x + x2 
dx 


/clx 

! + 0 + i)2 


sir- 

Vb^  1 + 


i=(fe 

a/3 


(x  -f 


*>» 


= -4=  tan"1  (x  + £)  i* 

Vo  V 3 

dx 


jy  _ 1 « = tan  1 (x  — !>• 

n ay  2 — 2x  + x3  y 


72.  cr’3  = 


(m  + nx)_dx 


a3  + z2 


« = - tan-1  - + | log  («3  + x3). 
^ a a 2 


254 


TRIGONOMETRIC  REDUCTION. 


135.  Trigonometric  Reduction.  — A very  slight  ac- 
quaintance with  Trigonometry  will  enable  the  student  to 
solve  the  following  examples  easily.  After  a simple 
trigonometric  reduction,  the  integrals  are  written  out  by 
inspection. 

1.  dy  = tan3  xdx. 

Here  y = J tan3  xdx  — j* (sec2  x — 1)  tan  xdx 


sec2  x tan  xdx  — tan  xdx] 


= l tan2  x — log  sec  x.  [See  (9)  of  Art.  132.] 

2.  dy  — tan4  xdx. 

y = tan3  x — tan  x + x.  [(15)  of  Art.  132.] 

3.  dy  = tan5  xdx. 

y = } tan4  x — I tan2  x + log  sec  x. 

4.  dy  = cot3  xdx.  y = — \ cot2  x — log  sin  x. 

5.  dy  = cot4  xdx.  y — — £ cot3  x + cot  x -f-  x. 

6.  dy  — cot5  xdx. 

y = — i cot4  x + cot2  x + log  sin  x. 


7.  dy  = sin3  xdx. 


Here  y = j £ (3  sin  x — sin  3x)  dx ; (from  Trig.), 


.'.  y = cos  3x  — J cos  x ; 
or  V — \ cosS  x — cos  x* 

8.  dy  — cos3  xdx.  y — sin  x — ^ sin3  x. 


* By  employing  different  methods,  we  often  obtain  integrals  of  the  same 
expression  which  are  different  in  form,  and  which  sometimes  appear  at  first  sight 
not  tc  agree.  On  examination,  however,  they  will  always  he  found  to  differ  only 
by  some  constant  ; otherwise  they  could  not  have  the  same  differential. 


EXAMPLES. 


255 


9.  dy  = cos4  xdx. 

Here  y — j* ( cos2  xf  dx  — f \ [1  -f  2 cos  2a; 

+ cos2  2a;]  dx  [See  (17)  of  Art.  132.] 
= + i sin  2a;  -f  \ J cos2  2xd  ( 2x ) 

•-  \x  4-  \ sin  2a;  + f [x  + f sin  4a;]  ; [by  (17)  of  Art.  132.1 
.*.  y — ^ sin  4a;  -f  \ sin  2x  + \x. 


10. 

dy 

— sin4  xdx. 

y = 

A sin  Ax  ~ 1 

sin  2a;  -f  fa;. 

11. 

dy 

= sin5  xdx. 

y = 

— COS  X + | cos3  x — \ cos5 

12. 

dy 

= cos5  xdx. 

y = 

sin  x — | sin3 

x \ sin5  x. 

13. 

dy 

= sin3  x cos3 

xdx. 

y - 

= cos  6x 

— Ti 

: cos  2a;.  (From  Ex.  7.) 

14. 

dy 

dx 

~ sin  x cos3 

X 

y = sec2 

x + log  tan  x. 

15. 

dy 

=r  sin3  x cos2 

xdx. 

y = \ cos3 

x — f cos3  X. 

16. 

dy 

= cos5  x sin5 

xdx. 

COS6  X 

y = -~2r 

'a- 

- f cos2  x + \ 

cos4  x). 

17. 

dy 

= sin6  x cos3 

xdx 

— (sin6  x — sin8  x)  cos  xdx. 

y =.  sin7  x 

(I  — i sin2  x). 

18. 

dy 

sin5  x , 
= — 5-  dx. 

cos2  X 

n i-2 

cos2 

x + cos4  x)  sin  xdx 

- s 

- j 

cos2  X 

= sec  x + 2 cos  x — f cos3  x. 

19.  dy  = cos7  xdx. 

y = sin  x — sin3  x + f sin5  x — \ sin7  x. 


CHAPTER  II 


INTEGRATION  OF  RATIONAL  FRACTIONS. 

136.  Rational  Fractions. — A fraction  whose  terms  in 
volve  only  positive  and  integral  powers  of  the  variable  is 
called  a -rational  fraction.  Its  general  form  is 

axm  + bxm~l  + cxm~2  + etc.  ...  I . . 

a'xn  + b'xn~l  + dxn~-  + etc.  ...  I” 

in  which  m and  n are  positive  integers,  and  a,  b,  . . ., 
a',  b',  . . . are  constants. 

When  m is  > or  = n,  (1)  may,  by  common  division,  be 
reduced  to  the  sum  of  an  integral  algebraic  expression,  and 
a fraction  whose  denominator  will  be  the  same  as  that  of  (1) 
and  whose  numerator  will  be  at  least  one  degree  lower  than 
the  denominator.  For  example, 

ofi  _ 2 2a-2  + x 

Z3  — X2  + X l — 1 X Xs  — X2  + x + 1 

The  former  part  can  be  integrated  by  the  method  of  the 
preceding  chapter ; the  fractional  part  may  be  integrated  by 
decomposing  it  into  a series  of  partial  fractions,  each  of 
which  can  be  integrated  separately.  There  are  three  cases, 
which  will  be  examined  separately. 

137.  Case  I.  — When  the  denominator  can  be  re- 
solved into  n real  and  unequal  factors  of  the  first 
degree 

f {.'x  \ dec 

For  brevity,  let  - v‘ . , - denote  the  rational  fraction 

J f(x) 

whose  integral  is  required,  and  let  (x—a)  (x — l)  . . . ( x—l ) 
be  the  n unequal  factors  of  the  denominator.  Assume 


DECOMPOSITION  OF  FRACTIONS. 


25? 


/M 

<p{x) 


A 


+ 


B 


+ 


C 


x — b x — c ’ ' * x — l 


L-,<  a) 


where  A,  B,  C,  etc.,  are  constants  whose  values  are  x>  be 
determined. 

Clearing  (1)  of  fractions,  by  multiplying  each  numerator 
by  all  the  denominators  except  its  own,  we  have 

f(x)  = A (x—b)(x—  e).. . (x—l)+B(x—a)[x—c).'.{x—T) 

-f  etc.  + L (x—a)  (x—b) . . . [x—h),  (2) 

which  is  an  identical  equation  of  the  ( n — l)th  degree.  To 
find  A,  B,  C,  etc.,  we  may  perform  the  operations  indicated 
in  (2),  equate  the  coefficients  of  the  like  powers  of  x by  the 
principle  of  indeterminate  coefficients  in  Algebra,  and  solve 
the  n resulting  equations.  The  values  of  A,  B,  C,  etc.,  thus 
determined,  being  substituted  ir.  (1)  and  the  factor  dx  intro- 
duced, each  term  may  be  easily  integrated  by  known 
methods. 

In  practice,  however,  in  this  first  case,  there  is  a simpler 
method  of  finding  the  values  of  A,  B , etc.,  depending  upon 
the  fact  that  (2)  is  true  for  every  value  of  x.  If  in  (2)  we 
make  x = a,  all  the  terms  in  the  second  member  will  re- 
duce to  0,  except  the  first,  and  we  shall  have 

f (a)  = A (a  — b)  (a  — c) ...  (a  — l), 


or 


A = 


m 


(a  — b)  (a  — c)  . . . (a 


_ _ ZM. 

1)  <p'  (a) 


In  the  same  way,  making  x — b,  all  the  terms  of  (2) 
disappear  except  the  second,  giving  us 

f{b)  — B (b  — a)  (b-c)...(b-I), 

f(b) 


G? 


B = 


(b  — a)  {b  — c) 


f(b) 

( b — l ) o {b) 


258 


CASE  1. 


Or,  in  general,  the  value  of  L is  determined  in  any  one 
of  the  terms,  by  substituting  for  x the  corresponding 

root  l of  <b  ( x ) in  the  expression  ; i.  e,,  L = • 

V 1 <P  (®)  Q (l) 


EXAMPLES. 

...  , , (x2  + 1 )dx 

l.  Integrate  iy  = ^+\x,  + ^ + g. 

In  this  example,  the  roots  * of  the  denominator  are  found 
by  Algebra  to  be  — 1,  — 2,  — 3. 

z3  + 6z2  -f-  llz  -f-  6 = (x  + 1)  (x  4-  2)  (x  + 3). 
Assume 

z2  + 1 A . B C 


+ 


+ 


(1) 


x 3 + 6z2  4-  llz  4-6  x 4-  1 24-2  24-3 

«\  z3  4- 1 = .4  (z  4-  2)  (z  4-  3)  4-  B (z  4 1)  (z  4-  3) 

4-  0 {x  1)  (x  4-  2). 

Making  x = — 1,  we  have  2 = 2 A,  r.  A = h 
“ x = — 2,  “ « 5 = — By  B = — 5. 

“ 2 = - 3,  « “ 10  = 2(7,  6’  = 5. 

Substituting  these  values  of  A,  B,  C,  in  (1),  and  multi- 


plying by  dx,  we  have 

y = f. 


(z2  4-  1)  dx 


x 3 4-  6z3  4-  liar  4-  6 

/dx  , P dx 


e£z 


4*  1 1/24*2  «/  c 4-  3 

y = log  (2  4-  1)  — 5 log  (2  4-2)4-5  log  (2  4-  3) 

(2  4-  1)  (2  4-  3)5 

^+2)3 


= log 


* If  the  factors  of  the  denominator  are  not  easily  seen,  put  it  equal  to  0,  and 
solve  the  equation  for  at;  the  first  root  may  be  found  by  trial,  x minus  each  of  the 
Several  roots  in  turn  will  be  the  factors.  (See  Aigebra.l 


DECOMPOSITION  OF  FRACTIONS. 


259 


2.  Integrate  dy  — 


adx 
x2  — a 2 
x 


s'  = ll0«?T i = logV/^‘ 

( o’2  _L_  9.^  rlnr 

3.  Integrate  dy  — 


V-\  log 


x4  — 5x2+  4 
x2 — x — 2 


4.  Integrate  cly 


x2-\-x—% 

(5x  + 1)  dx 
x2  + x — 2 


5.  Integrate  dy  — 


y — log  ( x — l)2  (x  + 2)s. 
dx 


a2  — b2x 2 


y 


1 , (a  + 

“ 2 ab  l0§  \«  - bx)' 


138.  Case  II. — When  the  denominator  can  be  re- 
solved into  n real  and  equal  factors  of  the  first  degree- 


Let  the  denominator  of  the  rational  fraction 


/0*0 
(p  (x) 


con- 


tain n factors,  each  equal  to  x — a. 

Assume 

m = -I  + + 1 

<p  (x)  (x  — a)n  (x  — a)"-1  ( x — a)K~* 

+ ' * ' {x  — a) 

Clearing  (1)  of  fractions  by  multiplying  each  term  by  the 
least  common  multiple  of  the  denominators,  we  have 

f ( x ) = A + B (x  — a)  + C(x  — a)2 

+ • • • L (x—a)n~l,  (2) 


(1) 


which  is  an  identical  equation  of  the  ( n — l)th  degree.  To 
find  the  values  of  A,  B,  C,  etc.,  we  equate  the  coefficients 
of  the  like  powers  of  x,  as  in  the  preceding  Article,  and 
solve  the  n resulting  equations.  The  values  of  A,  B,  C. \ 


260 


CASE  IT. 


etc.,  thus  determined,  being  substituted  in  (1),  and  the 
factor  dx  introduced,  each  term  may  be  easily  integrated 
by  known  methods. 

In  this  case  we  cannot  find  the  values  of  A,  B,  C,  etc.,  by  the  second 
method  used  in  Case  I,  but  have  to  employ  the  first.  When  both 
equal  and  unequal  factors,  however,  occur  in  the  denominator,  both 
methods  may  be  combined  to  advantage. 


EXAMPLES. 


1.  Integrate  dy  = 

k 2 — 3a;2  A B C 

Assume  ^ + 2^3  - (*  + 2)3  + {x  + 2)2  + x + 2 

2 — 3x2  = A + Bz  + 2B+  Cx2  + ^Cx  -f  4(7. 

A + 2B  + 4(7  = 2.  (2) 

B + 4(7  = 0.  (3) 

C=  -3.  (4) 

Solving  (2),  (3),  and  (4),  we  get 

A — — 10,  B = 12,  C = — 3. 

Substituting  these  values  of  A,  B,  and  C in  (1),  and 
multiplying  by  dx,  we  have 


(2  — dx2)  dx  _ 

~lx+W~  ~ ~ 


3 dx 

(x  + 2)3  ‘r  (x  + 2 f ~ x + 2 


10 dx  12 dx 

+ 


_ P (2  — 3x2)  dx 

^ ~ J (x  + 2)3 

5 12 


(x  + 2)2  x + 2 


— 3 log  ( x A 2\ 


_ t , 7 (cP  + x)  dx 

*'  Inte6rate  d»  = 


DECOMPOSITION  OF  FRACTIONS. 


261 


Assume 

^t±JL — - — — [ — 1 (1) 

(x-2)2(x-  1)  (x  — 2)2  ^ (x  — 2)  ^ x — 1 V J 
x2  + x = A (x  — 1)  + B (x  — 2)  (x  — 1) 

+ G{x-  2)2.  (2) 

Here  we  may  use  the  second  method  of  Case  I,  as  follows : 


Making  x — 2,  we  find  A — 6. 

“ x = 1,  “ “ B — 2. 

Substituting  in  (2)  for  A and  C theii  values,  and  making 
e = 0,  we  find 

0 = — 6 + 25  + 8 ; B — — 1. 


Substituting  in  (1),  and  multiplying  by  dx,  we  have 
P (x2  + x)  dx 

' = J l 


(x  — 2 f (x  — 1) 


P 6dx  P dx  P 2 dx 

J (x  _ 2)2  “ J + J J 


2 dx 


{X  - 2)2 


x — 2 


— log  (x  — 2)  + 2 log  {X  — l). 


6 , , (x-iy 

•••  ^=-Jvr2  + bgp^2r 

. T i j (3a;  — 1 )dx 
3.  Integrate  dy  = _ - 


8 


y=mx-s 

x2  — 4x  + 3 
9x 


+ 3 log  (a  — 3) 


4-  Integrate  dy  = ^ _ 6x2-+ 


dx. 


y = log  [ x (x  — 3)2]a. 


262 


CASE  III. 


139.  Case  III. — When  some  of  the  simple  factors 
of  the  denominator  are  imaginary. 

The  methods  given  in  Arts.  137  and  138  apply  to  the 
case  of  imaginary,  as  well  as  to  real  factors ; but  as  the  cor- 
responding partial  fractions  appear  in  this  case  under  an 
imaginary  form,  it  is  desirable  to  give  an  investigation  in 
which  the  coefficients  are  all  real.  Since  the  denominator 
is  real,  if  it  contains  imaginary  factors,  they  must  enter  in 
pairs  ; that  is,  for  every  factor  of  the  form  xffia  + JV—  1, 
there  must  be  another  factor  of  the  form  x ± a — by/  — 1,* 
otherwise  the  product  of  the  factors  would  not  be  real. 
Every  pair  of  conjugate  imaginary  factors  of  this  form  gives 
a real  quadratic  factor  of  the  form  ( x a)2  + b2. 

Let  the  denominator  contain  n real  and  equal  quadratic 
factors.  Assume 


f (x)  Ax  + B Cx  + D 

(x)  ~ [(x  ± aY  + ^2]n  + [(x  ± aY  + &]*-1 

Kx  4-  L 

+ (x  ± a)2  + V2  vl) 

If  we  clear  (1)  of  fractions  by  multiplying  each  term  by 
the  least  common  multiple  of  the  denominators,  we  shall 
have  an  identical  equation  of  the  (2 n — 1/A  degree. 
Equating  the  coefficients  of  the  like  powers  of  x,  as  in  the 
two  preceding  Articles,  and  solving  the  2 n resulting  equa- 
tions, we  find  the  values  of  A,  B,  C,  etc.  Substituting 
these  values  in  (1),  and  introducing  the  factor  dx,  we  have 
a series  of  partial  fractions,  the  general  form  of  each  being 

(Ax  + B)  dx 
\Jx~±a)2  4-  &]n’ 

in  which  n is  an  integer. 

To  integrate  this  expression,  put  x±a  = z;  .*.  x=  z^a, 


• Called  conjugate  Imaginary  factors. 


EXAMPLES. 


263 


dx  = dz,  ( x ± a)2  = z2.  Substituting  these  values,  we 
have. 


/ 


(Ax  + 5)  dx 
[( x ± af  + 62]' 


z T Aa  + B)  dz 
(z2  + b2)n 


r Azdz  r(B  - F Aa)  dz 

J (z2  •+-  ~W)n  J (z2  + b2)n 


A r A’dz 

2(n-  1)  (z2  + &2)"-1  + J (z2  + 62)*» 

(when  ,4'  = B Aa) ; 


«o  that  the  proposed  integral  is  found  to  depend  on  the 
integral  of  this  last  expression;  and  it  will  be  shown  in 
Art.  151  that  this  integral  may  be  made  to  depend  finallv 

Up0n  f tan_1  l * (Art- 132»  3.) 


EXAM  PLES. 


xdx 

1.  Integrate  dy  = -j- 

The  factors  of  the  denominator  are 

(x  — 1)  and  (x2  + x + 1). 
therefore  assume, 


+ 


Bx  + C 


Xs  — 1 X — 1 X1  -f  X + 1 

\ x = Ax3  + Ax  -f-  A -f-  Bx 2 + Ch-  — Bx  — (7. 

•.  d 4-  £ = 0;  yl  — B (J  — 1 ; ^-C=0. 

■4  = 1;  £ = - 1 ; <7  = *• 

_ P xdx  _ P dx  P (x  — 1)  dx 

^ J & — i J * x — i ^ + z + 1 


a) 


v 


264 


EXAMPLES. 


= l log  (*  — 1)  — i ; (2) 

(by  changing  the  form  of  the  denominator.) 

Put  x + £ = z,  then  x — 1 = z — f,  and  dx  = Jz,  and 
the  second  term  of  (2)  becomes 

_ i Ag  ~ t)  dz  _ _ i f zdz  tif  dz 

* J 22+|  s</z2  + f + 2t/22+f. 

= — £ log  (z2  + |)  H — tan-1  — (Art.  132,  3.) 


= — £ log  + * + 1)  + — f tan“l 

V3 


2x  + 1 

V3  ' 


(by  restoring  the  value  of  z). 


Substituting  in  (2),  we  have,  * 

v = i P°g  ix  — !)  — h log  + * + 1) 

+ V3tan 

A/3  J 

2.  Integrate  dy  = 

To  find  the  factors  of  the  denominator,  put  it  = 0 and 
solve  with  respect  to  a;2 ; thus, 


ad  -f-  ad  — 2 = 0,  or  ad  -f  ad  = 2. 


Assume 

Hence 


x2  = — £ ± f = 1 or  — 2. 

/.  ad  + a;2  — 2 = (ad  — 1)  (x2  + 2). 

x2  A B Cx  + D 

ad  + :c2  — 2 — z~+l  + x — 1 + ~f  + 2 ' 

x2  = A (x  — 1)  (ad  + 2)  + B (x  + 1)  ( x 2 -f  2) 
■+•  (Cir  .D)  (a;  — 1)  (a:  1). 


(1) 


(2) 


EXAMPLES 


36a 


AVe  may  equate  the  coefficients  of  the  like  powers  of  x , 
to  find  the  values  of  A,  B,  C,  D,  or  proceed  as  follows  : 

Making  x = — 1,  we  find  A = — -g-. 

“ x = 1,  “ “ B = f 

Substituting  these  values  of  M and  Z?  in  (2)  and  equating 
the  coefficients  of  a;3  and  a:2,  we  have 


6(7=  0 
(7=0 


and 

and 


6Z>  = 4; 
£ = §• 


••  9 = - fl  + + x* 


dx 

T» 


u . a/2  . .a? 


log(jxi)“  + ^tan  1^‘ 


1.  % = 

2.  cfy  = 

3.  dy  = 

4.  dy  = 

5.  dy  — 

6.  dy  = 


EXAMPLES. 

(a;  — 1)  dx 


x2  + 6x 

+ 8 

-f-  o)  dx 

Xs  + X2  - 

— 2x 

(3x2  - 

- 1)  dx 

+ 

t-H 

1 

1) 

adx 

x2  + bx 

(2  + 3x 

— 4x2) 

dx 

4x 

— X3 

y 

- 

(5x  — 

2)  dx 

, (*  + 4)- 

'“Nrrijr 

y — log  -f— — ” j* 

3 xi(x  + 2)4 

y = log  (x3  — a;). 

S'  = ,0«  (vTi)1, 


y = log  [a£  (2  + *)*  (2  — x)\ 
(x  + 2)3 


a:3  + 6X2  + 8a: 

12 


y = log  - 


x*  (x  + 4)^ 


266 


EXAMPLES. 


7.  dy 

8.  dy  = 

9.  dy  = 

10.  dy  = 

11.  dy  = 


(< a 3 + bx2)  dx 
a2x  — x? 


y = a log  x — --~-b  log  (a*  — x2). 


(3a;  — 5)  dx 
x2  — Qx  + 8 

x2dx 


(x  — a)2  (x  + a) 


V = 


2 (x  — a) 

( x 3 + x2  + 2)  dx 
x (x  — l)2  (x  + l)2 

3 + x 


, {x  — 4)? 

y ■=.  log  —• 

(x  - 2)i 


+ log  (x  — a)f  [x  + a)$' 


y — — 


+ log 


2 (x2  - 1)  ' ~6  (x  _ (X  + i)| 
x2dx 


(x  + 2)2  (x  + 1) 

4 

V = ^f~2  + l0g  ( x + 1)* 

12-  *°(.  + ^+'y 

3 +*(£# 


y = — 


x + 3 


13.  dy  = 
y 25  (x  — 2) 


dx 


(x  - 2)2  (a-  + 3)2 
1 


- tIt  log  (a;  - 2)  - 


1 


25  (x  4 3) 

+ rfr  l°g  (x  •+  3) 


14.  dy  = 


(x2  — 4x  4-  3)  dx 
x 3 — 6a;2  4-  9a; 

y — log  [x  (z  — 3f]k 


EXAMPLES. 


26? 


15.  dy  = 


x2dx 


(x  + 2)a  (a:  + 4)a 

5a:  4-  12  . (x  + 4\* 

8 + log  (x  + 2/  ‘ 


y = - 


x2  + 6a;  + 8 


'rflr 

1&  * = 5rnjF+ir 


y = \ tan  1 x + log 


17.  dy  = 


dx 


x 3 — a?  + 2a;  — 2 
(*  - 1)* 


y = log 


(x2  + 2)> 


(x2  + l)1 
(x  -f  1)1 


1 . ® 
— — tan  1 —=■ 

3a/2 


(x  - 2)$ 
(a?  + 1)* 


18  d..  - + g) - 
y — f tan-1  a;  + log 

* * = ^T^TT 

t 9a;2  + 9a;  — 128  , 

20,  dy  — x?  — ox2  + Bx~+9  C 

7 (« - 3)17  , 5 

y = log  + x~^~s' 


21.  = 


2xdx 


(x2  + 1)  (x2  + 3) 

i lxi  + 

y = lo«  UTs)  ' 


268 


EXAMPLES, 


an  , ( x 5 — 1)  dx 

22-  dy=y—^L]r- 


y = x + log  [(«  + 2)?  (z 


S3,  dy  = 


zafz 


(z  + 1)  {x  + 2)  (z2  -f-  ]) 


" log 


0 


(z  + 2)t  (z2  + l)-; 

(z  + 1)* 


a)*]. 

J f A tan-1 1 


CHAPTER  III. 


INTEGRATION  OF  IRRATIONAL  FUNCTIONS  BY 
RATIONALIZATION. 

140.  Rationalization. — When  an  irrational  function, 
which  does  not  belong  to  one  of  the  known  elementary 
forms,  is  to  be  integrated,  we  endeavor  to  rationalize  it ; 
that  is,  to  transform  it  into  an  equivalent  rational  function 
of  another  variable,  by  suitable  substitutions,  and  integrate 
the  resulting  functions  by  known  methods. 

141.  Function  containing  only  Monomial  Surds.— 

When  the  function  contains  only  monomial  surds,  it  can 
be  rationalized  by  substituting  a new  variable  with  an 
exponent  equal  to  the  least  common  multiple  of  all  the 
denominators  of  the  fractional  exponents  in  the  given 
function. 

For  example,  let  the  expression  be  of  the  form, 


..  _ gm'nee  • %n  — rfi'mce  ; %c  — tfnnc't  ; = gmnce  _ 


a'x°  -f  i’xfi 


Put 


rjQ  r/}7lTlC6  • 


e 


dx  = mncezmrux~xdz. 


Hence 


qJ  gmne'e  y zmnce‘ 


mncetfnnc*-1dz ; 


which  is  evidently  rational. 


270 


FUNCTIONS  CONTAINING  BIN^LL^m  RDS. 


1.  Integrate  dy  — 

Put 


X'i 


- dx. 


1 — X3 

x = z6; 

then  x*  = z8,  xi  = z2,  and  dx  = 6zsdz; 

(1  — z3)  6z5dz 


(1) 


= 6( 


•••  <*»= 

2s  + S4  — Z3  + Zs  — 2 + 1 — . ) az. 

1 + z/ 


Integrating  by  known  methods,  and  replacing  z by  ite 
ralue,  we  have 


y = 6 


r*  7 5 2 1 1 

x « z*  ira  X*  xs  i , 

-7  + T - T + -o  - ¥ + **  - lo§  0 + xi 


2.  Integrate  dy  — 


4 3 2 

3^  cte 


>} 


2x * — x t 


y = — 18  ^jr  + ^ + 4a£  + 16z^  + 32  log(2— x*) 


142.  Functions  containing  only  Binomial  Surda 
of  the  First  Degree. — When  the  function  involves  no 

m 

surd  except  one  of  the  form  (a  + bx)n,  it  can  be  rationalized 
as  in  the  last  Article,  by  treating  a + bx  as  the  variable. 
And  therefore  can  be  integrated. 

For  example,  let  the  expression  be  of  the  form, 

j xndx 

y ~ vT+ 1/ 


where  n is  a positive  integer. 

Put  a + bx  = z2 ; 


then 


x — 


z2  — a 


b ’ 


and 


FUNCTIONS  CONTAINING  BINOMIAL  SURDS.  271 


, xndx  2 ( z 2 — a)n  dz 

dy  = TitT*  = — ** — 

This  may  be  expanded  by  the  Binomial  Theorem,  and 
each  term  integrated  separately.  It  is  also  evident  that  the 
expression 

xndx 

p 

(< a + lx)* 

can  be  integrated  by  the  same  substitution. 


1.  Integrate  dy  = 


dx 


xVl  + x 

Put  1 + x = z2;  then  dx  = 2 zdz  and  x = £ — 
dx  2 dz 


dy  = 


*V  1 +X  z2  — 1 
dz 
+ ' 


= — ^ - — ^-r  (Art.  137) ; 
z — 1 z + 1 


or 


, 2—1  ! Vl  + x — 1 

y = ■<*,— t = l0«  vrriTT 


2.  Integrate  dy  = 
Put 


x?dx 


(1  + 4x)» 

(1  + Ac)  = 22 ; 


then  iz  = f,  (1  + 4z)5  = *. 


dy  = 


1 (22  _ 1)3 

128  24 


-i  r q i I 


y = t? 


128 


272  FUNCTIONS  CONTAINING  TRINOMIAL  SURDS. 


or 


y~ 


128 


-3(1+4#- 


(1  + 4®)*  + 3(l  + 4®)$_ 


143.  Functions  of  the  Form  — '—  } , where  n 

is  a Positive  Integer.  ( a + bx2)- 

Put  a + bx2  = z2 ; 


then  xdx  = ~ , x2  = 
a:2"*1  dx 


zi  — a 


x?n  = 


(z2  — a )* 


bn 


(z2  — a)n  dz 


(a  + bx2)* 


which  may  be  expanded  by  the  Binomial  Theorem,  and  each 
term  integrated  separately.  It  is  also  evident  that  the 
expression 

x2n+1  dx 


(. a + bx2)* 

can  be  integrated  by  the  same  substitution. 

1.  Integrate  dy  = — ===• 

Put  1 — x2  — z2 ; then  xdx  = — zdz,  x2  = 1 — t8. 
x?dx 


dy  — 


2.  Integrate  dy  = 


Vl  — a;2 

y = i*3  — 2 

o?dx 


— — (1  — z2)  dz. 

x2)*  — (1  — a?)*. 
(2a  + 3cr2) 


= i(l 


(a  + ca:2)v 


y = 


3c2  (a  + ca:2)^ 


144.  Functions  containing  only  Trinomial  Surds 
of  the  Form  V a + bx  ± x2. 

There  are  two  cases,  according  as  x2  is  + or  — . 


FUNCTIONS  CONTAINING  TRINOMIAL  SURDS.  273 


Case  I. — When  x2  is  +. 
Assume 


\/a  + bx-\-x2  = z — 


then 


a + bx  = z2  — 2 zx ; 


x — 


zi  — a 


b -f-  2z 


Av  — 2 (z*  + iz  + G)  dz 

(b  + 2 zf 


and  Va-{-bx-\-x2  = z — 


24  — a 
b + 2z 


z2  + bz  + a 
2 z + b 


The  values  of  x,  clx,  and  a / a + bx  + x2  being  expressed 
in  rational  terms  of  z,  the  transformed  function  will  be 
rational,  and  may  therefore  be  integrated  and  the  z replaced 
by  its  value  V a + bx  + x2  -f  x. 


Case  II. — When  x2  is  — . 

Let  a and  0 be  the  two  roots  of  the  equation 
x2  — bx  — a — 0 ; 

then  we  have  x2  — bx  — a — (x  — a)  (x  — 0)\ 
a + bx  — x2  = — (x  — ci)  (x  — 0) 
= (x  — «)  (0  — a-). 

Assume  V a + bx  — x2  = V — «)  {0  — x) 

= (x  — a)  z ; 

(x  — «)  ( 0 — x)  {x  — a)2  z2, 
or  (0  — x)  — (x  — «)  z2  \ 

az2  + 0 


whence, 


and 


x = 


dx  = 2 
V a + bx  — x2  = ^ 


z2  + 1 ’ 

(ct  — 0)  z dz 

~W+T)^’ 

az2  + 0 
z 2 + 1 
(0  - a)  Z 

Z2  + 1 


-4 


274 


EXAMPLES. 


The  values  of  x,  clx,  y/  a + bx  — x2,  being  expressed  in 
rational  terms  of  z,  the  transformed  function  will  be  rational 


1.  Integrate  dy  = 


dx 


+ bx  + x1 


Assume  V a + bx  + x2  = 2 — x\ 

then,  as  in  Case  I,  we  have 


a bx  — z2  — 2 zx ; 


x = 


b + 2k 


dx  = 


y/ a + bx  -f  x2  = 


dy  = 


2 (z2  + bz  + a)  dz 
(6  + 2 zf 

z2  + bz  + a 
2 z + b 

2 (z2  + bz  + a)  dz  x (2z  + b) 
(b  -t-  2 z)2  x ( z 2 + bz  + a) 

2 dz  dz 


b + 2z 


2 


+ 2 


v = /r^-  = 1°s(l  + 2) 


+ 2 


= log 

If  b = 0,  we  have 

y = f 


+ x + V a + bx  + x2 


']■ 


dx 


V a + x? 
and  if  a = 1,  we  have 

dx 


y 


= / 


Vi  + 


— log  (x  + V a 4-  14) 


= log  ( X + \/l  + X2). 


EXAMPLES. 


275 


2dz 

Had  we  integrated  the  expression  - — — without  dividing  both 

0 4-  &Z 

terms  of  it  by  2,  we  would  have  found  for  the  integral  the  following-. 
y = log  (b  + 2z)  = log  [b  + 2x  + 2^  a + bx  + x2\,  which  difl'ers  from  the 
above  integral  only  by  the  term,  log  2,  which  is  a constant.  (See 
Note  to  Art.  135  ) 

2.  Integrate  dy  = — 

V«  + bx  — x 2 

Let  cc  and  8 be  the  roots  of  x2  — lx  — a = 0 ; then,  as  in 
Case  II.  we  have 


's/ a + bx  — x2  = (a 


(i 6 — x)  = (x  — cc)  z2 ; 


cc)  (/3  — x)  = (x  — cc)  z. 

ccz2  + (3 


x - 


, _ 2(cc—(3)  zdz  _ 

~ {z2  + 1)2  ’ 


dy  = 

y = / 


Va  + bx  — x2  = 
2 («  — 8)  zdz  ( 2 2 + 1) 


z2  + 1 

(8  — cc)z 
z2  + 1 

2 dz 


(z2  + l)2  (0 

dx 


cc)  z 


= = - 2 f 


3.  Integrate  cfo/  = 


+ bx  — ar 

= — 2 tan-1  z = — 2 tan-1 
dx 


1 + z2’ 

dz 

1 + z2 


\ x — a 


C\/l  + 2 + 22 


Assume  \Zl+x  + x2  — z—x,  and  we  have,  as  in  Case  I, 


x = 


dy  = 


z2  — 1 7 2 (z2  + z + 1)  dz 

r + * = ~ a + &>. — 
vrrvT?  = ^+^- 

2 (z2  + z + 1)  dz  (1  + 2 z)  (1  + 2z)  _ 

(1  + 2 z)2  (z2  + z + 1)  (z2  — 1)  “ 

dz  dz 

z — 1 z + 1 


2dz 


(Art.  137). 


276 


BINOMIAL  DIFFERENTIALS. 


dx 


Vl  + x + cc2 


log  = log  ^l±vrg+g 

2 + 1 * + 1 + \/\  -f  X + a? 

. Zx 

lo°*  - ■ ■ — — • 

° 2 + x + 2^/i  + x + x1 


145.  Binomial  Differentials. — Expressions  of  the  form 
dy  — xm  (a  + bxn)p  dx, 

in  which  m,  n,  p denote  any  numbers,  positive,  negative,  or 
fractional,  are  called  binomial  differentials. 

This  expression  can  always  be  reduced  to  another,  in 
which  m and  n are  integers  and  n positive. 

1st.  For  if  m and  n are  fractional,  and  the  binomial  of  the 
form 

ar  ^ (a  -f-  bx^)p  dx, 

we  may  substitute  for  x another  variable  whose  exponent  is 
equal  to  the  least  common  multiple  of  the  denominators  of 
the  exponents  of  x,  as  in  Art.  141.  We  shall  then  have  an 
expression  in  which  the  exponents  are  whole  numbers. 
Thus,  if  we  put  x = z6,  we  have 

x~  * (a  + bx*f  dx  = 6z~3  (a  -f  bz3Y  dz, 

in  which  the  exponents  of  z are  whole  numbers,  and  the 
exponent  of  z within  the  parenthesis  is  positive. 

2d.  If  n be  negative,  or  the  binomial  of  the  form 

xm  (a  + bx~nY  dx, 

we  may  put  x — - , and  obtain 
z 

xm  (a  + bz~nY  dx  — — z~m~2  ( a + bznY  dz, 

in  which  the  exponents  of  z are  whole  numbers,  and  the 
one  within  the  parenthesis  is  positive. 


CONDITIONS  FOR  RATIONALIZATION. 


277 


3d.  If  £ be  in  both  terms,  or  the  binomial  is  of  the  form 
xm  (ax?  + bxn)v  dx, 

we  may  take  xf  out  of  the  parenthesis,  and  we  shall  have 

xm+pt  lxn-ty  fa, 

in  which  only  one  of  the  terms  within  the  parenthesis  con- 
tains the  variable. 


146.  The  Conditions  under  which  the  General 
Form  _ 


dy  = xm  (a  -Y  bxn)q  dx, 


can  be  rationalized,  any  or  all  of  the  exponents  being  frac- 
tional. 


(1.)  Assume 

a + hxn  = 

p 

Then 

(a  + bxn)q  — 

ZP. 

(1) 

Also 

x — 

lzq  — 

V ~b~)  * 

m 

id 

xm  = 

w- 

(2) 

dx  — 

1 *=> 

^ ^ 

T 

CT* 

53-1^ 

(3) 

Multiplying  (1),  (2),  and  (3)  together,  we  have 

P ^±l_i 

dy  = x7?  (a  + bxn)idx  = ^zP+?_'(2 — (4) 


an  expression  which  is  rational  when 
or  0. 


m + 1 . • , 

is  an  integer, 

n ° 


(2.)  Assume  a -f-  bxn  — zq xn. 

Then  xf1  = a iyfl  — b)~\ 


(1) 


278 


CONDITIONS  OF  INTEG R ABILITY. 


x — ah  (zq  — b)~k  (2) 

m m 

af  = a « (z?  — b)~ ”,  (3) 

cfe  3=  — | (z«  — 2?-1  C?z.  (4) 


Multiplying  (1)  by  b,  adding  a,  and  taking 
have 


V 

Q 


power,  we 


V P _p 

(a  + bxn)q  = aq  (z9  — b)  q zP. 


(5) 


Multiplying  (3),  (4),  and  (5)  together,  we  have 


xf"  (a  + bxn)q  dx 


U 

n 


(”+■ 
r \n  q 


p+L)  -('511+p+i') 

9 •'(*_  J)  v " ? ’zr+t-'dz, 


an  expression  which  is  rational  when  m b - is  an  in- 
teger, or  0.*  U y 

Therefore  there  are  two  cases  in  which  the  general  bino- 
mial differential  can  be  rationalized : 


1st.  When  the  exponent  of  the  variable  without  the 
parenthesis  increased  by  unity,  is  exactly  divisible  by 
the  exponent  of  the  variable  within  the  parenthesis. 

2d.  mien  the  fraction  thus  formed,  increased  by 
the  exponent  of  the  parenthesis,  is  an  integer. 

Rem.— These  two  cases  are  called  the  conditions  of  integrahility  of 
binomial  differentials,!  and  when  either  of  them  is  fulfilled,  the  inte- 
gration may  be  effected.  If,  in  the  former  case,  — ~ — 1 is  a positive 

771  + 1 77 

integer  or  0,  or  in  the  latter  case, + - + 1 is  a negative  integer 


* The  student  will  observe  that  Art.  143  is  a particular  case  of  this  Article,  re- 
sulting from  making  m an  odd  positive  integer,  and  n = 2. 

+ These  are  the  only  cases  of  the  general  form  which,  in  the  present  state  of 
analysis,  can  be  made  rational.  When  neither  of  these  conditions  is  satisfied,  the 
V 

expression,  if  - be  a fractional  index,  is,  in  general,  incapable  of  integration  in  » 
finite  number  of  terms. 


EXAMPLES. 


279 


or  0,  the  binomial  (2?  — a)  or  (2?  — b)  will  have  a positive  integral  ex- 
ponent, and  hence  can  he  expanded  by  the  Binomial  Theorem,  and 

each  term  integrated  separately.  But  if,  in  the  former  case,  — — - — 1 

n 

is  a negative  integer,  or  in  the  latter,  — f-  - + 1 is  a positive  inte- 

ger, the  exponent  of  the  binomial  (p  — b)  will  be  negative,  and  the 
form  will  be  reduced  to  a rational  fraction  whose  denominator  is  a 
binomial,  and  hence  the  integration  may  be  performed  by  means  of 
Chapter  II.  But  as  the  integration  by  this  method  usually  gives  com- 
plicated results,  it  is  expedient  generally  not  to  rationalize  in  such 
cases,  but  to  integrate  by  the  reduction  formula  given  in  the  next 
Chapter. 

1.  Integrate  dy  = x5  (a  -f  x2)*  dx. 
in  - 4-1 

Here 1 = 2,  a positive  integer,  and  therefore  it 


ran  be  integrated  by  the  first  method. 


Let 

( a + x2) 

- = Z3. 

Then 

(a  + x2)z 

= Z. 

(1) 

X6 

= (z3  — 

Clf, 

x5  dx 

II 

*«!co 

1 

- a)2  z2dz. 

(2) 

Multiplying  (1)  and  (2) 

together. 

, we  have 

dy  - 

x5  (a  + x2)  3 dx 

= K23  - 

- a)2  z3dz. 

y = 

- J*  ( z 9 — 2 £a  + a2z3)  dz 

3 /z10 
“ 2\10  ' 

2 az7 

+ t) 

— 

its  (a  + xT) 3 — 

3 ci , 

■ y («  + 

2,i  3a2 

x2)*  + 8 

(a  + Z2)s. 

AT  l 

2.  Integrate  dy  = r = ar4(l  -f  x2)~^  dx. 

xf  (1  + x2Y 

TT  m - f 1 p , — Iff  1 -, 

n q 2 2 

ative  integer,  and  hence  it  can  be  integrated  by  the  secoud 
method 


280 


EXAMPLES. 


Let  (1  + £2)  ~ zV. 

Then  z2  = (z3  - l)-». 

X = (z3  — l)-4 ; 

Z-*=(Z3-1)3  (1) 

(1  + X3)  = 1 + (z3  - 1)^ 

= Z2  ( z 3 — 1)-*. 

(1  + a«)-i  = z~'(z3-  l)i  (2) 

dx  = — (z3  — 1)~’  zdz.  (3) 

Multiplying  (1),  (2),  (3)  together,  we  have 
dy  = x~*  (1  4-  x3)~i  dx  = — (z2  — 1)  <iz. 


y = — J*  ( z 3 — 1)  c Iz  = z — \£ 

_ (1  + z2)*  h (1  + z2)1  _ (1  4 x3)* 


x? 


dx3 


(2s»  - Ip. 


1.  dy  = 


(Art.  14L) 


2.  dy  = 


EXAMPLES. 

(2a:^  — 3a:*)  dx 

5x& 

y = A®*  - f®*- 

x^  — 2x* 


dx. 


1 + x* 

y — %x*  — 2x  — £.r*  + 3a^  + 2a^  — 6z$  — 62^ 
+ 6 log  (23  + 1)  + 6 tarrtcs'. 


? dy  = 
y = 

+- 1908  [*** 


2xi  — 32;*  , 

5 rdx. 

3Z*  4-  x? 

12  (%x*  — \xi  -f-  \^x^  — 9a:^) 

— £.r*  + 3zi — + 81^ — 243  log  (V*  + 3)1 


EXAMPLES. 


281 


, 1 + X*  , 

dy  = rdx- 

1 + X* 


ii  a t l l 

„„  \a;T*  a;3  a;^  xs  t x*  / in 

^ “ 12  |lT  ¥ 7 X ¥ 3“  + i^°g(1  + a:3) 


- f.  f: log  (**  - y + *)  + 3 tan- 

2f  L \a;^  + 2^T^  + 1/ 

5.  dy  = — (Art.  142.) 


-(rS)]| 


(i  + xy 


y = 2 

6.  dy  = 


Vl  + a;  + 


Vl  + 


Tl] 


dx 


%V  a + bx 


2 . V a + bx  — V~a 

y = T^ vs1 — 


7.  ty  = 


(1  + x)i  + (1  + *)* 
y — 2 tan-1  (1  + a;)^. 

dy  — 4 (a:  + V x + 3 + v^a:  + 3)  $r. 
y — 2 (a;  + 3)2 — 12  (a: + 3)  + f (a;  + 3)-  + 3 (a:  + 3)t 

(Art.  143.) 


, aftfa; 

* = vrX? 


10.  = 

11.  dy  = 


y = |(1  + a?)f  - | (1  + afi)*  + (1  + z2)*- 

a^^  3a^  + 2 


(1  + a*)* 

x?dx 

(1  + a*)*' 


y = - 


3 (1  + a;2)* 
.c2  + 2 


y a/1  + a? 


282 


EXAMPLES. 


12.  dy  == 


dx 


(Art.  144.) 


13.  dy  = 


Vl  + x + x2 
y = log  (l  + 2 x + 2V 1 + x + V). 
(See  Art.  144,  Ex.  1.) 
dx 


14.  dy  = 

15.  dy  = 


V x2  — x — 1 
y — log  ( 2x  — 1 + 2 V x2  — x — l). 
dx 


V '2  — x — x 2 
dx 

Vi  + x — 7? 


y = — 2 tan-1 


V— 

V x + 2 


y = — 2 tan-1 


£ — £a/5  — a: 
a:  — | — 4V5 


Va2  + i^z2 

^Assume  ^ + x2  = z — x,  etc.'j 


y = | log  (&r  + V^T^2> 

l?.  dy  = — =•  y = - log 

zV  «2  + a 


Vo2  (•  — 


-) 


or 


_i|°g(: 


lx 

a + \/ a2  -f- 

&r  ./ 


is.  iy  = (**  + *)*& 


y = log  {x  + 1 + \/2£  + x2)  — 


+ V%X  -f  X2 


EXAMPLES. 


283 


19.  dy  = 

V = 

20.  dy  = 

y 

21.  dy  = 

22.  dy  = 


dx 


(1  + x2)  Vl  — x2 

X tan-  (-^L). 

V2  Wl  — xV 


adx 


\^2ax  + x 2 

a log  {x  + a + V 2 ax  + x2). 
dx 


y 


= £ log  ( 2x  -f-  V 4x2  — 7). 


V 4X2  — 7 

(Compare  with  Ex.  16.) 

2 dx  . . 

V^T^'  <,  = log(a:+S+V}*  + 4 

(Compare  with  Ex.  20.) 
cd  (2  + 3x2)?  dx.  (Art.  146.) 


27 


'(2  + 3.r2)^ 


23.  dy 

y 

24.  dy  — x*  (a  + bx2)?  dx. 


-|(2  + 3^  + 


f(2  + 3^)i]- 


y 

25.  dy 

y 

26.  dy 

y 

27.  dy 


= (a  + dx2)?  ( 


hbx2  — 2a\ 


\ 35b2  / 

x3  (a  + bx2)?  dx. 

lip  — 2d)  (a  + bx2)*. 
x3  (a  — x?)~?  dx. 

— i (a  — x2)^  (2a  -f-  x3). 
dx 

~ S*  + 


X2  (1  + X2)? 

28.  dy  = a (1  + x2)~?  dx. 


y = 


-^)i(^  + 24 


y = 


ax 

(1  + 


284 


EXAMPLES. 


29.  dy  : 

30.  dy  : 

31.  dy  : 

32.  dy  : 

33.  dy  : 

in  which  z 

34.  dy  ■ 

35.  dy  : 


ar4  (1  — 2 x*)~i  dx. 


(1  + a;2)*5  a^a:. 


y = 


_ (3a2  - 2)  (1  + z2)1 


15 


= ar2  (a  + a^)-*  dx. 

(3a;3  + 2a) 

y 2a2a;  (a  + a^)s 

= ar3  (a2  + x2)i  dx. 

V = A («2  + a?)*  (4a:2  — 3a2). 

= a^  (a  + to2)^  <7a:. 

_ 3 z5  / z6  az3  a2\  < 

^ 2^s vn  t + 5/  ; 

= (a  + bx2)  3. 

(a  + Ja:)®  <£c 

x 

y — | (a  + + 2a  (a  + 

s i Va  + — V a 

+ «*  log  -p- 

Va  + aa:4-  v a 

= (a2  + a:2)^  dx. 

y = ~r  | log  [2  + (a2  + z2)-]- 


CHAPTER  IV. 


INTEGRATION  BY  SUCCESSIVE  REDUCTIONS. 


147.  Formulae  of  Reduction. — When  a binomial  dif- 
ferential satisfies  either  of  the  conditions  of  integrability,  it 
can  be  rationalized  and  integrated,  as  in  the  last  chapter. 
But,  instead  of  rationalizing  the  integral  directly,  it  may 
be  reduced  to  others  of  a simpler  kind,  and  finally  be  made 
to  depend  upon  forms  whose  integrals  are  fundamental,  or 
have  already  been  determined.  This  method  is  called 
integration  by  successive  reduction,  and  is  the  process 
which  in  practice  is  generally  the  most  convenient.  It  is 
effected  by  formulae  of  reduction.  These  formulae  are 
obtained  by  applying  another,  known  as  the  formula  for 
integration  by  parts,  and  which  is  deduced  directly  as 
follows  : 

Since  d (- uv ) = udv  + vdu,  (Art.  16) 


we  have 


uv 


a formula  in  which  the  integral  of  udv  depends  upon 
that  of  vdu. 


148.  To  find  a formula  for  diminishing  the  expo- 
nent of  x without  the  parenthesis  by  the  exponent 
of  x within,  in  the  general  binomial  form 


C\ xm  (a  + bxll)p  dx. 


286 


SPECIAL  FORMULAS  OF  REDUCTION. 


Let  y = J* xm  ( a + bxn)p  dx  - --  J* udv 

= uv  — f vdu  ; (1) 

and  put  dv  = xn~l  ( a + bxn)p  dx  and  u — 

(a  + bxu)p+1 


Then 


v — 


nb  (p  + 1)  ’ 
and  du  = (m  — n + 1)  xm~ndx. 

Substituting  these  values  of  u,  v,  du,  dv,  in  (1),  we  have 

xm--n+\  (a  _|_  J)Xny+ 1 


y = xm  (a  -f-  bx")p  dx  — 


nb  (p  + 1) 

fr-°(a  + h?)>«ax.  (2) 

This  formula  diminishes  the  exponent  m by  n as  was 
desired,  but  it  increases  the  exponent  p by  1,  which  is 
generally  an  objection.  We  must  therefore  change  the 
last  term  in  (2)  into  an  expression  in  which  p shall  not  be 
increased. 

Now  xm~n  (a  + bx")p+1  = xm~n  (a  + bx'Y  ( a + bxn) 

= axm~n  ( a + bxn)p  -f  bxm  ( a + baf‘)p ; 
which  in  (2)  gives 


V 


— j* xm  (a  + bx“)p  dx  = 


r-"+1  (a  + bxf')p+l 
nb  (p  + 1) 


-Z~Pn++i)  fxm{a  + dx- 

Transposing  the  last  term  to  the  first  member  and  redu- 
cing, we  have 


SPECIAL  FORMULAE  OF  REDUCTION. 


287 


/np  -f  m + 1 
\ n (p  + 1) 


— ^ J xm  ( a + bxn)p  dx  — 


xm~nJrx(a  + bxn)p+1 


nb  (p  + 1) 

m n a J xm~n  ( a + bx^Y  dx. 


nb  (p  + 1) 
Therefore  we  have 


y = J ' xm  (a  + bxnY  dx 

xm-,'+l(a-\-btfiy+x  — ( m — n-\-  l)a J xm~n(a-\-bxn)pdx 


b (np  + m + 1) 
which  is  the  formula  required. 


149.  To  find  a formula  for  increasing  the  exponent 
of  x without  the  parenthesis  by  the  exponent  of  x 
within,  in  the  general  binomial  form 

y — f x~m  (a  + bxtl)p  dx. 

Clearing  (A)  of  fractions,  transposing  the  first  membei 
to  the  second,  and  the  last  term  of  the  second  to  the  first, 
and  dividing  by  (m  — n + 1)  a,  we  have 

J* xm~n  (a  + bxnY  dx 

xm~n+](a  + bxn)p+l — b(np  + rn  + 1 ) f x?1  (a  + bxnydx 

— y (i) 

a (m  — n + 1) 

Writing  — m for  rn  — n,  and  therefore  — m + n for  m, 
(1)  becomes 

y = f x~m  (a  + bxt'Y  dx 


x~m+\a  + bxf'Y+1 + b(m—np—n— 1)  J x~m  -n(a  + bxn)vdx 


— a (m  — 1) 

which  is  the  formula  required,. 


288 


SPECIAL  FORMULAE  OF  REDUCTION. 


150.  To  find  a formula  for  diminishing  the  expo 
nent  of  the  parenthesis  by  1,  in  the  general  bino- 
mial form 

y — J xm  ( a + bxn)p  dx. 

J* xm  {a  -f  ixn)p  dx  = J* xm  ( a -f  bxn)rr~l  (a  + bxn)  dx 

— a J xm  {a  bxn)p~x  dx  + b J xm+n  ( a + bxn)p~x  dx.  (1) 

By  formula  [A),  we  have  for  the  last  term  of  (1),  by  writ- 
ing tn  + n for  rn  and  p — 1 for  p, 

J' xm+n  ( a + bzn)p~l  dx 

xm+1  (a  + bxn)p  — (m  + 1)  a i xm  ( a + bxn)p~x  dx 
b [n  (p  — 1)  + in  -f  n + 1] 
which  in  (1)  gives 

y = f x"1  (a  + bxn)p  dx  = a f* xm  (a  + baf')p~x  dx 

at"+1  ( a + bx^y  — (in  + 1)  a C xm  (a  + bxf)p~x  dx 

*J 


+ 


( np  -f-  m -f-  1) 


Therefore,  uniting  the  first  and  third  terms  of  the  second 
member,  we  have 

y — f xm  (a  -f-  bxny  dx 
x?l  rX  ( a -f  bxn)p  + anp  ( xrn  ( a -f  bxn)p-1  dx 

ey 


np  -j-  m + 1 

which  is  the  formula  required. 


(C) 


SPECIAL  FORMULA  OF  REDUCTION. 


289 


151.  To  find  a formula  for  increasing  the  exponent 
of  the  parenthesis  by  1,  in  the  general  binomial  form 

y = J xm  ( a + bxn)~p  tlx. 

By  transposing  and  reducing  ( C ),  as  we  did  (H)  to  find 
(B),  we  have 

J* xm  ( a + bxn)p~1  dx 

xm+1  ( a bxn)p  — (np  + m + 1)  f xm  [a  + bx")p  dx 

= -(1) 

— anp 

Writing  — p for  p — 1,  and  therefore  — p + 1 for  p, 
(1)  becomes 

y — f xm  (a  + bxn)~p  dx 


xm+1(a  + bof)~p+l  — (m  + n + l—np)  f xm(a  + bxn)-pJr'dx 


an  ( p — 1) 

which  is  the  required  formula. 


;C Dt 


Remark. — A careful  examination  of  the  process  of  reduction  by 
these  formulae,  will  give  a clearer  insight  into  the  method  than  can  he 
given  by  any  general  rules.  We  therefore  proceed  at  once  to  exam- 
ples for  illustration,  and  shall  then  leave  it  to  the  industry  and  inge- 
nuity of  the  student  to  apply  the  method  to  the  different  cases  that 
he  may  meet  with. 


EXAMPLES. 


1.  Integrate 


, xmdx 

dy  — -7== 


V a2  — 


Here  y = f xm  ( a 2 — x2)~?  dx, 

a form  which  corresponds  to 


290 


EXAMPLES. 


We  see  that  by  applying  formula  (A)  we  may  diminish 
m by  2,  and  by  continued  applications  of  this  formula,  we 
can  reduce  in  to  0 or  1 according  as  it  is  even  or  odd,  so 
that  the  integral  will  finally  depend  upon 


or 


/ 

/ 


dx  . x 

— sin*1  - , when  m is  even  ; 


Va2  — x2 

xdx 
V a2  — x 2 


= — (a2  — x2)i,  when  m is  odd. 


Making  m — in,  a — a2,  i — — 1,  n = 2,  p = — 
we  have  from  formula  ( A ), 

y = J "* xm  ( a 2 — x2)~i  dx 

xm~ 2+1  ( a 2 — x2)^  — a2  (in  — 2 + 1)  ^ xm~2(a2  — x2)~?  dx 


— [2  ( — £)  + OT  + 1| 


xfn~1  ( a 2 — x2)i 


m 


+ 


(in  — 1)  a2  j' xm~2  ( a 2 — x2)~?  dx 


in 


0) 


When  m — 2,  (1)  becomes 

r x2dx  x a2  . x 

!/  = JvS^=-2ia  X)  +2  ‘m  a 

When  in  = 3,  (1)  becomes 
y?dx 


= /: 


= — \x2  (a2  — x2)i  — fa2  (a2  — +) - 


\/  a 2 — x2 
— — l (a2  — .r2)v  ( x2  + 2a2). 

When  m — 6,  (1)  becomes,  by  applying  (+)  twice  in 
succession, 


EXAMPLES. 


291 


= / 


rMx 
V a2  — x2 


— (a2  — x2) 3 


re5  5a2x3  5a4x\  ha6 

.6  + (TT  + 4T4  / + 4^1 


sin-1  - 


(which  the  student  may  show.) 

xmdx 


2.  Integrate  dy  — 


V a2  + 


Here 


y 


= j* xm  (a2  + x2)~?  dx. 


Making  vi  = to,  a = a2,  b — 1,  n = 2,  p = — 
we  have  from  ( A ), 

y = J xm  (a2  + x2)~i  dx 

= r^+.^)*_(E-l )*  fx.-,(a,  + a?)-idx.  (1) 

m in  «./  v ' v ' 


By  continued  applications  of  this  formula,  the  integral 
will  finally  depend  on 


/ 

/ 


dx 

\/  a2  + x2 
xdx 

^ a2  + x2 


— log  ( x + V a2  -f  x2),  when  rn  is  even, 
= (a2  + x2)i,  when  to  is  odd. 


3.  Integrate 
Here 


dy  = 


dx 


xm  (a2  — «2)^ 

y = J* x~m  ( a 2 — a;2)-*  dx. 


from  which  we  see  that  by  applying  ( B ) we  may  increase  to 
by  2,  and  by  continued  applications  of  (£),  we  may  reduce 
to  to  0 or  1,  according  as  it  is  even  or  odd,  making  the 
integral  finally  depend  on  a known  form. 


i«i  g s 


292 


APPLICATIONS  OF  FORMULA. 


Making  m = to,  a = a%  b = — 1,  n = 2,  y = — 

( B ) gives  us 

y - J* x ~m  ( a 2 — z2)-^  dx 

x~mH  ( a 2 — x2p  — (to  + 1—2—1)  J' x~m+2(a2 — z2)  (fa 
— a2  (to  — 1) 

(a2  — z2)?  (to  — 2)  /'  m 

(to  — 1)  c2*”1-1  + (to  — 1)  a2t/  xm-2  (a2  _ ' 

When  to  = 2,  (1)  becomes 

dx  (a2  — x2)? . 


V 


= /- 


a;2  V a2  — z2 
(since  the  last  term  disappears.) 
When  to  = 3,  (1)  becomes 
dx 


alx 


y 


= /; 


:3  V d2  — X2 


{a2  — z2)*  1 P dx 

+ \ 


2a2x2 


x V a2  — z3 


V a2  — x2  1 a — V a2  — x2 

— * Incr  — 

2a3 


2a2z2  + l0S 


(Ex.  17  of  Art.  146.) 

■4.  Integrate  dy  = ( a 2 — z2)^  dx,  when  n is  odd. 

Here  we  see  that  by  applying  ( C ) we  may  diminish 

by  1,  and  by  continued  applications  of  (C)  we  can  reduce 

to  — making  the  integral  depend  finally  upon  a 
known  form. 

tl 

Making  to  = 0,  a = a2,  l = — 1,  n — 2,  p = ^ , ( C) 
gives  us 


y = f \a2  — z2)*  dx 
x (a2  — z2)2  + na2  j ( a 2 — z2)*-1  dx 


n + 1 


(1) 


APPLICATIONS  OF  FORMULA. 


293 


When  n ~ 1,  (1)  becomes 

y = f(a>  - *•)»  <fc  = + £ sin-1  ?■ 


5.  Integrate  dy  = 


dx 


Here 


(a2  — z2)* 
y = f (a2-  a:2)—"  ah', 


- , when  re  is  odd. 


from  which  we  see  that  by  applying  (Z))  we  may  increase 
the  exponent  - by  1,  and  by  continued  applications  of  {D) 

7Z 

we  can  reduce  --  to  — making  the  integral  depend 
£ 

finally  on  a known  form. 

Makii 
gives  us 


Making  m — 0,  a = a2,  b = — 1,  re  = 2,  p = ( D ) 

/V 


y = J* (a2  — a:2)  - dx 

(a2  — x2)  2 1 — (3  — re)  f ^ (a2  — z2)-^1  dx 


2 a2 


(l-i) 


+ 


re  — 3 p 
— 2)  a2  J i 


dx 


(re  — 2)  a2  (a2  — z2)t  1 ^ (a2  — z2) 

When  n — d,  (1)  becomes 


•M  « 


= / 


rfz 


(«2  _ (fl2  _ *2)1 

xmdx 


6.  Integrate  dy  = — 

-V  2az  — z2 

Here  y = f xm  (2 ax  — z2)-*  dx  — J* xm~?(2 a — x)~^dx, 
which  may  be  reduced  by  (A)  to  a known  form. 


294 


APPLICATIONS  OF  FORMULAE. 


Making  m — m — a = 2a,  b = — 1,  n = 1,  p = — 
(A)  gives  us 

/x™dx 
V2  ax  — x1 

x™-?  (2a  — x)i  — 2 a ( m — \)  / xm~i  (2a  — z)--  dx 


— m 


x * /r 7,  (2m  — 1)  a xm~ldx  .... 

— V2az  — z2  + i — / — (1) 

m m c * 


/s/2ax  — z2 


When  m = 2,  (1)  becomes 

/z2(?z  z + 3a  /jr — „ 

— = — y2az  — z2 

V 2az  — z2  * 


+ f«2  f— 
7 «/ 


c?z 


V2az 


_ — £ \/ 2az  — z2  + |a2  vers-1  - 

■).  i a 


7.  Integrate  dy  = — - 


x6dx 


V 1 — z2 


/z5  1*5  , 1-3-5  \ r- — -5  ,1-3.5.  , 

» = - («  + IT6^  + 2TT7C I)'/l-^+2T4-6sln_‘I- 

8.  Integrate  dy  = — g 

z4  Va  + fa? 

y = (-  ss  + Jl) v;FT^- 

9.  Integrate  dy  = (1  — z2)v  dx. 

y — \x  (1  — z2)^  + fz  (1  — z2)?  + | sin-1  %. 

T dx 

10.  Integrate  dy  = + -2y 


z 


z 


3 


L 0 GARITEMIC  FUNCTIONS. 


295 


11.  Integrate  dy  — 


x?dx 


\/2ax  — X 2 

y = — (^  + i ^ a + i ’ t^2)  ^ ^ax  — + 1 • |«3  vers'"1  ?• 


12-  Integrate  dy  — 


dx 


x?  V 1 — a£ 


y “ (l./4  + 2 • 4a:2)  V1 


1 • 3 


* — 271 log 


1 + Vl  — x2 


These  integrals  might  be  determined  by  one  or  other  of 
the  methods  of  Chapter  III,  but  the  process  of  integration 
by  reduction  leads  to  a result  more  convenient  and  better 
suited  in  most  cases  for  finding  the  definite  integrals.* 


LOGARITHMIC  FUNCTIONS. 

152.  Reduction  of  the  Form  ^ X (log  x)n  dx,  in 
which  X is  an  Algebraic  Function  of  x. 

Put  Xdx  - dv  and  log"  x = u. 

/clcc 
Xdx  and  du  = n log"-1  x — • 

Substituting  in  J udv  = uv  — J*  vdu,  (Art.  147) 

we  have  y = J X log"  xdx 

= log"  x J* Xdx  — J* n log"-1  x (Xdx) ; 

or  by  making  J (Xdx)  = X,, 

we  have  y = f X log"  xdx 

rX 

= Xi  log"  x — n J — 1 log"-1  xdx ; 


* For  a discussion  of  definiie  integrals,  see  Chap.  V. 


296 


EXAMPLES. 


which  diminishes  the  exponent  of  log  x by  1,  wherever  it 

is  possible  to  integrate  the  form  f Xdx.  By  continued 

applications  of  this  formula,  when  n is  a positive  integer, 
we  can  reduce  n to  0 so  that  the  integral  will  finally 
depend  on 


Sch. — A useful  case  of  this  general  form  is  that  in  which 
X = xm,  the  form  then  being 

y — / xm  log"  xdx ; 

and  the  formula  of  reduction  becomes 


by  means  of  which  the  final  integral,  when  n is  a positive 
integer,  becomes. 


EXAMPLES. 


1.  Integrate  dy  — xi  log2  xdx. 

Making  m = 4,  and  n = 2,  we  have 


Making  m = 4 and  n = 1,  we  have 


LOGARITHMIC  FUNCTIONS. 


297 


which  substituted  in  (1)  gives  us 


= J ar1  log1 


2 x dx  = 


a;5  log2  a:  2 hr5  log  x 1 a^\ 


5 5 


x5 


= jr  (\0g*X-%l0gX  + *). 


2.  Integrate  dy  = 


x log  a;  da; 

Vffl2  + x2 


Put 


then 


xdx 


\/  a2  x2 


= dv 


and  log  x — u ; 


dx 


v — V a2  + x2  and  du  = — 


rx  log  x dx  , „ , i . rVa2  + : 

••  , = ^vfe  = ( + 1 8I“^^ 

a2dx  r xdx 


dx 


= (°!  + ^ Iog  * - “ /vp+» 

= (a2  + a^)*  log  a;  + a log  ^ a2 + x?. 


3.  Integrate  dy 


(See  Ex.  17,  Art.  146.) 
log  x dx 


(1  + x)2 


153.  Reduction  of  J* 


1 -f  x 

xm  dx 
log"  X 


log  X — log  (1  + x). 


Put 

then 

and 


ar+1  ==  u, 


1 dx 


log"  x x 
du  = (m  + 1)  xm  dx 
1 


= dv; 


v = 


— {n  — 1)  log"-1  x 


298 


LOGARITHMIC  FUNCTIONS. 


pxP  dx  _ xm+1  i m + 1 p xm  dx 

" y J ]og"x  ( n — 1) log"-1  x (n — 1 )J  \ogn~lx 5 

by  means  of  which  the  final  integral,  when  n is  a positive 
integer,  becomes 

Pxm  dx 

J log  x’ 

beyond  which  the  reduction  cannot  be  carried,  for  when 
n — 1 the  formula  ceases  to  apply.  We  may,  however,  ex- 
press this  final  integral  in  a simpler  form ; thus, 

Put  z = xm+x ; 

then  dz  — (m  + 1)  xm  dx  and  log  z = (rn  + 1)  log  x. 

'‘xm  dx  _ P dz 
log  x ~ J log  z ’ 

an  expression  which,  simple  as  it  appears,  has  never  yet 
been  integrated,  except  by  series,  which  gives  only  an 
approximate  result. 

7*4  cI  t 

Ex.  1.  Integrate  dy  = • 

Here  m — 4 and  n = 2 ; therefore  the  formula  gives  us 


...  r> 

j i 


y 


= /: 


xi  dx 
log2  X 


a? 

log  x 


+ 5 


f 


x 4 dx 
log  x 


Put  z — x 6;  then  dz  = 5xidx  and  log  z = 5 log  x; 

dz 


therefore 


P xt  dx  _ P 
J log  x — J 


log  2 


Now  put  log  z = u;  then  z = eu  and  dz  = eud'<L 

P dz  _ Peudu 
J log  2 — J u 

PL  u2  us  , \du  , . , 

= y(1+“+2+5T3  + etc')  V (Art-  6S) 


EXPONENTIAL  FUNCTIONS. 


299 


- IV  (Ay 

= lo  g“  + * + ^+  ^ + etc* 


= log  (log  X5)  + logz5  + ^log2^  + 
.r4  dx 

°*  ^ «/  loff2  X 


+ etc, 


log2  x 
ofi 


log  x 


+ 5 


log  (log  X 5)  + log  Xs  + 


log2  X? 

— ¥~ 


log3  OP 

+ 2-32  + e c' 


(See  Strong’s  Calculus,  p.  392  ; also,  Young’s  Int.  Cal., 
pp.  52  and  53.) 


EXPONENTIAL  FUNCTIONS. 


154.  Reduction  of  the  Form  f amx  xn  dx. 

Put  amx  dx  = dv  and  x*  — u ; 

nmx 

then  v — 

m log 

amixn 


— , aud  du  — nof~l  dx. 

m log  a 


•••  y = f' 


amxxn  dx  . 


:/ 


amxxn~ 1 dx. 


m log  a m log  a < 

By  successive  applications  of  this  formula,  when  n is  a 
positive  integer,  it  can  be  finally  reduced  to  0,  and  the  in- 
tegral made  to  depend  on  f amx  dx  = 


amx 


m log  a 


Only  a very  few  of  the  logarithmic  and  exponential  func* 
tions  can  be  integrated  by  any  general  method  at  present 
known,  except  by  the  method  of  series,  which  furnishes 
only  an  approximation,  and  should  therefore  be  resorted  to 
only  when  exact  methods  fail. 


300 


EXPONENTIAL  FUNCTIONS. 


EXAMPLES. 


1.  Integrate  dy  = axx?  dx. 

Here  m = 1 and  n = ‘6\  therefore,  from  the  above  for- 
mula, we  have 


= / 


ax'j?dx 


ax7? 
log  a 


3 

log  a 


J ' ax$dx\  (by  repeating  the  process) 


axx?  3 / uxx2 
log  a log  a \log  a 


log  a 


J axxdx\ ; 


(by  repeating  the  process) 
axJ*  3 axx2  6 / axx  1 \ 

log  a log2  a log2  a \log  a log2  a / 

_ ax  / 3a;2  6a;  6 \ 

log  a \ log  a log2  a log3  a) 

2.  Integrate  dy  = xh^  dx. 


y = gOX 


/x3 

\a 


3a?  6a:  _ 6 \ 

a2  a3  av 


/ax 

— , when  m is  a posi- 
tive Integer. 

Put  x~m  dx  = dv,  ax  = u ; 

£ — 771+1 

then  v = and  du  = ax\dgadx. 

m — 1 


/axdx  _ i 

xm  (j m — 


ax  log  a 

1)  xm~x  m 


axdx 


xm' 


-i  j 


by  means  of  which  the  final  integral  becomes 

raxdx 


TRIGONOMETRIC  FUNCTIONS. 


301 


which  does  not  admit  of  integration  in  finite  terms,  but 
may  be  expressed  in  a series,  and  each  term  integrated  sepa- 
rately. (See  Lacroix,  Caleul  Integral,  Vol.  II,  p.  91.) 

Ex.  1.  Integrate  dy  = “T' 

By  the  formula  just  found,  we  have 


TRIGONOMETRIC  FUNCTIONS. 

156.  Cases  in  which.  sinm  6 eosn  Odd  is  immediately 
Integrable. — The  value  of  this  integral  can  be  found  im- 
mediately when  either  ill  or  n,  or  both,  are  ocld  positive 
integers ; and  also  when  m + n is  an  even  negative 
integer. 

1st.  Let  m = 2r  + 1 ; then 
J* sin"1  0 cos"  0 dO  = J' (sin  0)2r+1  cos"  6 dO 


an  expression  in  which  the  binomial  (1  — cos20)r  can  be 
expanded,  and  each  term  integrated  immediately.  In  like 
manner,  if  the  exponent  of  cos  0 be  an  odd  integer,  we  may 
assume  n = 2r  + 1 , etc. 


- (Art.  62) 


302 


EXAMPLES. 


2d.  Let  m + n = — 2r ; then 
J sinm  0 cos"  d dd  — J tanm  0 (cos  6)n+m  dd 
= J tanm  0 (sec  6)2r  dd 
= J " tanm  *9(1  + tan2  0)'-1  d • tan  d, 


each  term  of  which,  after  expansion,  can  be  immediately 
integrated. 


EXAMPLES. 

1.  dy  = sin20  cos30  dO. 

Here  y = J sin20  cos30  dd 


2. 


= f sin20  (1  — sin2  0)^- sin  d 
— £ sin30  — \ sin50. 


dy 


sin2  d 
cos4  d 


dd. 


Here  y — / — ~ dd  = C tan2  d sec2  d dd 
J J cos4  d J 

= | tan3  d. 

3.  dy  — sin3  d cos4  d dd.  y = — | cos5  d + \ cos7  6 

4.  dy  = sin5  d cos5  d dd. 

y — — iV  cos6  d (sin4  d + \ sin2  d + |). 

5.  dy  — sin3  d cos7  d dd.  y = cos10  d — £ coss  d. 

ct  T1  ^ f) 

6.  dy  — — — dd.  y = 4 tan3  0 + 4 tan5  0. 

J COS6  0 V A 1 6 

„ n __  dd 

y — sin  0 cos5  0 


FORMULA  OF  REDUCTION. 


303 


Here  «=  /'( 

J J tan  9 cos2  6 J 


1 + tan2  0)2  sec2  9 dO 


tan  9 cos2  9 O tan  9 

= log  (tan  9)  + tan2  0 + J tan4  0. 
dO 


8.  dy  = - 3 B 
sin' 2 0 cos2  0 


Let 


x - - tan  6; 


1 3/ 

then  cos  0 = — , sin  6 — — - , 

Vl  + x2  Vl  + a? 


and 


^ = rr?’ 


...  i = /•  .» , = /< i 

v nr\Q'J  ft  ^ 


4-  a:2)  dx 


sin2  6 cos2'  0 
2 


= fa:*  — i _ j 
3 x*  3 


tan  2 (9  — 


a;2 

2 


tan2  6 


9.  iy  = ^dB. 
3 cos5  9 

10.  dy  = — 

^ cos6  6 


y ~ \ tan4  9. 

y = tan  0 4-  § tan3  0 4 £ tan5  6 


157.  Formulae  of  Reduction  for 

J* sinm  9 cos“  9 (19. 

When  neither  of  the  above  mentioned  conditions  as  to  m 
and  n is  fulfilled,  the  integration  of  this  expression  can  be 
obtained  only  by  aid  of  successive  reduction. 

We  might  produce  formulae  for  reducing  the  expression 
sin"‘  9 cos"  9 directly  ;*  but,  as  it  would  carry  us  beyond 
the  limits  of  this  book,  we  prefer  to  effect  the  integration 
by  transforming  the  given  expression  into  an  equivalent 
algebraic  form,  and  then  reducing  by  one  or  more  of  the 


* See  Price,  Lacroix,  Williamson,  Todhunter,  Courtenay,  etc. 


304 


EXAMPLES. 


formulae  (A),  (B),  ( C ),  (B).  Thus,  put  sin  0 = x,  then 
sinm  0 — xm,  cos  0 = (1  — x 2)'“',  cos”  0 = (1  — x2)t>,  and 
dO  = (1  — a:2)~2  (/a;. 


y — J* sinm  0 cos"  OdO  = xm  (1  — x2)  ~ dx. 
or  we  may  put  cos  0 — x,  and  get 

//»  m— 1 

sinm  0 cos"  Odd  = J — a?  (1  — x2)  • dx ; 


either  of  which  may  be  reduced  by  the  above  formulae. 

This  process  will  always  effect  the  integration  when  m 
and  n are  either  positive  or  negative  integers,  and  often 
when  they  are  fractions.  The  method  is  exhibited  by  the 
following  examples. 


EXAMPLES. 

1.  dy  = sin6  Odd. 

Put  sin  0 = x, 

then  dO  — (1  — x2)~*  dx. 

y = f*  sin6  OdO J x*  (1  — x2)~i  dx 

hr5  5Z3  3-5x\/1  ,.1  3-5  . , 

= - (e  + O + zpA  (1  - * + 2^  s,n' * 

(by  Ex.  7,  Art.  151): 

cos  0 . . . . ,5  . , 5-3  . 5-3 

= — (sin5  0 + ^ sin3  0 + ^ sm  0)  + „ . „ 0. 


2-4-6 


2. 


dy  = 


dO 

sin5  0 
sin  0 = x, 


dO  = (1  — a*)-*  i*. 


Put 

then 


INTEGRATION 


305 


= = /*"  <4  - **>'*  ix- 


1-3  . 1 + Vl  — 


/I  1-3  \ /- t 1-3  . 

= -fe  + ^Vl-^-8^l0g 

(by  Ex.  12,  Art.  151); 
cos  0/1  3 \ 1-3. 

= ~ ~ ure  + wV + 5U log  tln  ie- 

i ■ i 1 + cos  0 . sin  0 . 

(since  — log  — = log  — ■ = log  tan  \9.) 


sin  0 
3.  dy  — sin4  Odd. 
cos  0 


1 + cos  0 


V — 4—  (sin3  0 + f sin  0)  + § 9. 

(See  Ex.  10,  Art.  135.) 


y 


4.  dy  = cos4  9d9. 
sin  9 cos3  9 


+ | sin  6 cos  9 + f 0. 


(See  Ex.  9,  Art.  135.) 


158.  Integration  of  sinTO  0 cosn  0 d9  in  terms  of 
the  sines  and  cosines  of  the  multiple  arcs,  when  m 
and  n are  positive  integers. 

The  above  integrations  have  been  effected  in  terms  of  the 
powers  of  the  trigonometric  functions.  When  m and  n 
are  positive  integers,  the  integration  may  be  effected  with- 
out introducing  any  powers  of  the  trigonometric  functions 
by  converting  the  powers  of  sines,  cosines,  etc.,  into  the 
sines  and  cosines  of  multiple  arcs,  before  the  integration  is 
performed.  The  numerical  results  obtained  by  this  pro- 
cess are  more  easily  calculated  than  from  the  powers. 

Three  transformations  can  always  be  made  by  the  use  of 
the  three  trigonometric  formulas. 


306 


EXAMPLES. 


( 1 •)  sin  a sin  b — \ cos  (a  — b)  — \ cos  ( a + b). 

(2.)  sin  a cos  b = \ sin  (a  + b)  + \ sin  ( a — b). 

(3.)  cos  a cos  b = \ cos  (a  + b)  + £ cos  (a  — b). 


EXAMPLES- 

1.  dy  — sin3  0 cos2  Odd. 


sin  0 (sin  0 cos  0)2 

sin  0 (£-  sin  20)2 

[by  (2)] 

£ sin  0 (sin2  20) 

, . . /I  — cos  40\ 

i sm  0 (-  j 

[by  (1)] 

= £ sin  6 — ^ sin  0 cos  40 
= sin  0 — £ (•£  sin  50  — £ sin  30) 

I>y  (2)] 

= ■£■  sin  0 — sin  50  + -fa  sin  30 


,\  y = J* sin3  0 cos2  Odd 
= f \\  sin  Odd  — ^ sin  50dd  + sin  3 ddd) 


= — -J-  cos  0 + -gV  cos  50  — cos  30 

2.  dy  — sin3  0 cos3  Odd. 

y = — cos  20  + cos  60. 

3 cfy  = sin3  Odd. 

y = yV  cos  30  — £ cos  0. 

4.  = cos3  0fZ0. 

y — ^ sin  30  + | sin  0. 


FORMULA  OF  REDUCTION. 


3J? 


159.  Reduction  of  the  Form 

J xn  cos  ax  dx. 


Put 

and 

tliei 

and 


=f 


xn  cos  ax  dx 


u = xn, 


dv  = cos  ax  dx ; 
du  — nxn~l  dx. 


v — - sin  ax. 
a 


- x"  sm  ax 

a a 


— - fi sir 

a u 


sm  ax  dx. 


Again,  put 

u = x”-1, 

and 

dv  = sin  axdx; 

then 

du  = (n  — 1)  2 dx. 

and 

1 

v = cos  ax. 

a 

J* 3f~l  sin  ax  dx  = 

— - af*-1  cos  ax 
a 

-f-  — / xn~2  cos  ax  dz. 

a a 


\ y = J* xT  cos  ax  dx  = ^ x"  sin  ax 

— - (_  1 xn~l  cos  ax  + - / z*~2  cos  ax  dx\ 

a\  a a a I 

xn~'x  (ax  sin  ax  4-  n cos  ax)  n hi  — 1)  /*  „ „ 

= 1 — 5 — - / af'~~  cos  ax  dx 

a i a 2 J 

The  formula  of  reduction  for  C xn  sin  a%dx  can  be 
obtained  in  like  manner. 

EXAMPLE. 

1.  dy  = z3  cos  x dx. 

y = & sin  x -j-  3a?  cos  x — (5x  sm  x — 6 cos  x. 


S08 


FORMULA  OF  REDUCTION. 


160.  Reduction  of  the  Form 


J ea* 

cos’1  x dx. 

Put 

u — 

COS”  X , 

and 

dv  = 

eaz  dx ; 

then 

du  = 

— n cos"-1  x sin  x dx, 

and 

v = 

e“ 

a 


/,  cos”  x 

e“cos  nxdx  — — ^ 

+ ^ J cos"-1  x sin  x dx.  (1) 


Again,  put 
and 
then 

and 


u 

dv 

du 

v 


= cos”-1  x sin  x, 

= e^dx-, 

= — («  — 1)  cos"-9  x sin8  x dx 

+ cos"  x dx, 

e“x 

TTT  — • 

a 


J e"  cos*-1  x sin  x dx 


1 i r 

=—eax  cos"-1  x sin  x / e^f  — (n  — 1)  cos*-2  a;  sin8  x 

a ad  L 

+ cos"  x]  dx 

= - e“  cos*-1  x sin  x + — — f e“  cos"-9 a;  dx 

a ad 

— ” J cos”  x dx.  (Since  sin8  x = 1 — cos2  x.) 

Substituting  in  (1),  and  transposing  and  solving  for 
/*“  cos"  x dx,  we  get 


y — J* e"  cos"  x dx  = 


e1  cos”-1  x (a  cos  x + n sin  x) 
cfi  -f-  n2 


p-  f eai  cos"-9  x dx ; 

a 8 + n 2 d 


(2) 


EXAMPLES. 


309 


which  diminishes  the  exponent  of  cos  x by  2.  By  con- 
tinued applications  of  this  formula,  we  can  reduce  n to  0 
or  1,  so  that  the  integral  will  depend  finally  on 


when  n is  even ; 


31 


/' 


e"  cos  xdx, 


when  n is  odd. 


(2)  gives  the  value  of  J* &1*  cos  x d x without  an  integra- 
tion, since  the  last  term  then  contains  the  factor  n — 1 
= 1 — 1 = 0,  and  therefore  that  term  disappears. 

The  reduction  of  J eax  sin” a;  dx  can  be  obtained  in  like 
manner. 


EXAMPLES. 


1. 


2. 


dy  — (F  cos  x dx. 


y = 

dy  = 

y = 


(a  cos  x -f  sm  x). 

a2  + 1 

cos2  x dx. 

eF  cos  x(a  cos  x 4-  2 sin  x) 
4 + a2 


+ 


2 

4 + a2 


ff. 

a 


161.  Integration  of  the  Forms 

f(x)  sin-1  x dx , f(x)  tan-1  x dx,  etc. 

Integrals  of  these  forms  must  be  determined  by  the 
formula  for  integration  by  parts  (Art.  147) ; the  method 
is  best  explained  by  examples. 


1.  dy 
Put 

then 


EXAM  PLES. 

sin-1  x dx. 
dv  = dx, 

and 

u = 

sin-1  x\ 

ft  — x 

and 

du  = 

dx 

Vl  — x2 

310 


EXAMPLES. 


n=J 


sin  ~lxdx 


= x sm  1 x 


r xdx 

j a/1  " 


Vi  — & 

= x sin-1  x + (1  — x2)k  - p. 


2.  dy  = 


Put 


x2  tan-1  x dx 


1 -f-  x2 


dv 


x2dx 


= dx 


and 

then 

and 


1 + x2 
u = tan-1  x ; 
v — x — tan-1  x, 

7 dx 

du  = 


dx 

1 + x?* 


x dx  tan-1  x dx 
+ X2  1 + x2  I 


1 + X2 

n 

/.  y — x tan-1  x — (tan-1  x)2  — J \i 

— x tan-1  x — (tan-1  x)2  — 1-  log  (1  + x 2)  + ^ (tan-1  x)s 
= x tan-1  x — \ (tan-1  x )3  — £ log  (1  + x2). 

3.  dy  — x2  sin-1  x dx. 
rc3  . 


y = — sin-1  x + ^ (x2  + 2)  Vl  — x2. 

O 


4.  dy  = sin-1  x - 


dx 


(1  - z2)* 

162.  Integration  of  dy  = 

y = L de 


y = i (sin-1  xf. 


de 

a + b cos  d 


a + b cos  9 


d9 


EXAMPLES. 


311 


p dd 

(a  + b)  cos2 1 + (a  — b)  sin2| 

J 7„ 

sec2  H dd 

= r * 

a + b + (a  — b)  tan2  s 


Q 

d-  tan  - 

rl 


a + b + {a  — b)  tan2  5 


(1) 


When  ay  b, 


tan-1 


'(!=»)* tan  ?"| 
_\a  + bJ  2J 


Vcd—W 

(by  Ex.  3,  Art.  132.) 

When  a < b,  we  have,  from  (1), 
dd 


y 


= /•_**_  = 3 r 

J a + b cos  0 J 


d-  tan  - 

zV 


& + a — (5  — a)  tan2 


6 


Vb2- 


Vb  + a + 'Vb  — a tan  - 
log £ 

e 


Vb  + a — y/b  — a tan  - 


The  integral  of 


(by  Ex.  5,  Art.  137). 
dd 


to  be 


a -f  b sin  d 


can  be  found  in  like  manner 


. 0 , d 

2 a sin  g + b cos  - 

tan-1 — , when  ay  b; 


(a2  b 2)-'  (a2  — d2)^  cos 


fcSI 


312 


EXAMPLES. 


and  = 


-i  log 


a tan  - + b — (J2  — a2)* 


anaUi+V-#)  *’ 


when  a < b. 

There  are  other  forms  which  can  be  integrated  by  the 
application  of  the  formula  for  integration  by  parts  (Art. 
147).  Those  which  we  have  given  are  among  the  most 
important,  and  which  occur  the  most  frequently  in  the 
practical  applications  of  the  Calculus.  The  student  who 
has  studied  the  preceding  formulae  carefully  should  find  no 
difficulty  in  applying  the  methods  to  the  solution  of  any 
expression  that  he  may  meet  with,  that  is  not  too  compli- 
cated. 

The  most  suitable  method  of  integration  in  every  case 
can  be  arrived  at  only  after  considerable  practice  and  famil- 
iarity with  the  processes  of  integration. 


EXAMPLES. 


1.  dy  = 

2.  dy  = 


oPdx 

Vl  — x 2 
x*dx 


y=  -*(a»  + 2)(l-rf)*. 


Vl  — 2® 

/Xs  1-3  \ r- 5 1-3  . , 

y = - (r  + j-id  Vl  ~ * + an  '*• 


8.  dy  = 


xd dx 

Vi  — 


'A 


(sfi  1-6  1-4-6  „ , 1-2-4-6X  R -J 

» = -(?  +jr7*  + 3^*  + T^i)'/T=*- 

4.  dy 


xhlx 


V®  + 

1 /.  , 4 ax1  3a\  , — —5-5 

i'  = i5j(3^--r  + -»)  '/a  + w- 


EXAMPLES. 


313 


5.  dy  = 


dx 


=-( 


1-5 


6Z6  + 4-6X4  + 2 • 4 • 6x2, 


1-3-5 


)Vl- 


1-3-5.  1 + Vl  — a? 

log 


2-4-6 


6.  dy  = 


dx 


cVl  + a? 


y * \Vl  + x + 1/ 

(See  Ex.  1,  Art.  142.) 

7.  dy  — {a2  — x2)*  e?x. 

y — \x  (a2  — x2)±  + a2x  (a2  — x2)* 


+ eS  ^ ~ x2)i  + a° sin_1 


5-3 


6-4-2 
5x2  — 2 


(1  + n*)i. 


8.  dy  = x3  (1  + a?)*  dx.  y — g 5 

9.  dy  = (1  — x2)*  rfx. 

2/  = ix  (1  — x2)t  + £x  (1  — x2)*  + 1 sin-1  x. 

7 dx 

10.  dy  = -s* 

(a  + to2)1 

_ / 1 2\ x 

V ~ \«  + bx2  + a)  3aVa~+b^ 

dx 


11.  dy  — 


(a  + bx2)§ 

r 1 


4 8 "|  x 

y [_(a+A^)2  + 3a  {a  + bx2)  + 3 a2_  5 aVa  + bx2' 
14 


314 


EXAMPLES . 


12.  dy 

y 

13.  dy 

y 


x*dx 

(i  _ 3?yf 

x?  — 2>x 

— - — — sin-1  x. 

^Vl  — x1 


xidx 


■\/  2 ax  — x 2 


/a:3  7: 

= “W+i: 


7a;2  , 7 -5a;  , 7-5-3  ,\  _ 

4-3a  + 4-3-2G!  +4-3-2flj 


! ax—xi 


. 7-5-3  a; 

+ 4^s“  Ters  5- 

14.  dy  = log  a:  dx.  y = x (log  a:  — 1). 

15.  dy  — 3?  log2  a;  y = fa:3  (log2  a;  — ■§•  log  a;  + f ). 

y — log  (log  x). 


dx 


x log  x 
x log3  x dx, 


| log3  x — log2  x + ^logx- fa;2. 


16.  dy 

17.  dy 

y 

% 4 qA 

18.  dy  — 7?  log  x dx.  y = '—  log  x — — • 

19.  dy 

20.  dy 

21.  dy 

y 


dx 

x log2  x 
log2  x dx 

3 * 

X* 

x4dx 
log3  x 

x5 


y 


1 

log  X 
2 


y (log2  x -f  4 log  x 4 8), 

x* 


5X5  25 

2 log2  x 2 log  x 2 


log  (log  a?) 


4 log  x?  -f-  ~ log2  sfi  + loS3  ^ 


EXAMPLES. 


315 


22.  dy  = tFxldx. 

23.  dy  — xaxdx. 


24.  dy  — x2exdx. 

x2dx 


25.  dy  — 

26.  dy  = ^ 


e? 

axdx 


y — ex  (x*— 4^+  12a:2 — 242  + 24). 

ax  I 1 \ 

^ — log  a \ log  a) 

y = ex  (x2  — 2x  + 2). 
y = — e~x  (x2  + 22  + 2). 


2>  = “ £ + * lo£a)  + (l0S  -T  + lo£  a‘x 

+ \ log2  a-X-  + etc.^* 


27.  dy  = 


(e2®  — 1)  dx 
e2®  + 1 


28.  dy  — e*  e?  dx. 

e®2  dx 


29.  dy 
.30.  dy  = 


(1  + x? 

( 1 + 22)  exdx 

(1  + x)2 


y — log  (e®  + e x). 


y — ee. 


V = 


e® 


1 + 2 


y 


=*(~) 

\1  + 2/ 


[Put  (1+2 ) = z ; then  x — z — 1,  dx  — dz,  etc.] 


0,  , sm5  0 dd  . . 

31.  dy  — - 5-5—  (Art.  156.) 

J cos2  9 y ' 

y = sec  0 + 2 cos  9 — \ cos3  0. 

32.  dy  = sin7  9 cos3  9 d9.  y = § sin7  0 — f sin7  6. 

00  , sin3  0 c7(9  „ s . _ i „ 

33.  dy  — - y — t cos?  ® — 2 cos7  0. 

cos7  0 

_ , , cos3  9 d9  n . i , . 7 - 

34.  dy  = ^ — • ?/  = 3 sin7  0 — f sin  7 0. 

sin7  0 


316 


EXAMPLES. 


35.  dy 

y 

36.  dy 

y 

3 7.  dy 
38.  dy 


sin5  0 dd 
cos2  0 

— — — — (sin4  0 + 4 sin2  0 — 8). 
3 cos  0 v ' 

dQ 


sin4  0 cos2  0 

1 4 cos  0 8 cos  0 

cos  0 sin3  0 3 sin3  0 3 sin  0 

dd 

sin*  0 cos*  0 

sin*  0 d0 
cos*  0 
dd 


sin4  0 cos4  0 


39.  dy 

40.  dy  = sin4  0 cos4  0 dd.  (Art.  157.) 

y 


y — 2 tan*  0 (1  + tan2  0). 
y = | tan*  0. 

3/  — 8 cot  20  — | cot3  20. 


(cos3  0 + \ cos  0)  — (sin3  0 + § sin  0) 


41.  dy 

42.  dy 


dd 

sin  0 cos2  0 
dd 

sin  0 cos4  0 


64 
+ Tlb^- 


y = sec  0 + log  tan  -• 


y = 5 — 4 a + log  tan 

^ 3 cos3  0 cos  0 ° 2 

43.  dy 

y 


— sin8  0 cos6  0 ^0. 


cos^  0 

— — (sin7  0 + ^ sin5  0 + ^ sin3  0 +-^  sin  0) 
14: 


+ W(COS50  + 4cos30  + Y«os0)  + 


768 


EXAMPLES. 


311 


44. 

45. 

46. 

47. 

48. 

49. 

oO. 

51. 

Put 

52. 

y = 


dy  = sin4  9 dd.  (Art.  158.) 
y — sin  4 0 — £ sin  29  + |0. 
dy  — cos4  9 d9. 

y — fa  sin  40  + £ sin  29  + f 0. 
dy  = sin6  0 d9. 

y — fa  ( — ^ sin  60  + § sin  40  — sin  20  + 100). 
dy  — x 4 sin  x dx.  (Art.  159.) 
y = — x*  cos  x + 4a,'3  sin  x + 12a?  cos  x 

— 24a;  sin  x — 24  cos  x. 
dy  =.  e"  sin2  a:  dx.  (Art.  160.) 

eax  sin  x , „ , 2eax 


y 


--  (a  sin  a;  — 2 cos  x)  -f- 


a (4  + a2) 


4 + a 2 
t/y  = ez  sin3  x dx. 

y — faex  (sin3  x + 3 cos3  a;  + 3 sin  x — 6 cos  x ). 

a sin  Ar  + /r  cos  kx 


dy  = e-"  sin  Ac  c£r.  y = 
dy  = gAggA  (Art.  161.) 


(«2  + /I’2)  A 


Vi  — £2 


= 


a;3  Ac 


a/i  — 


— and  w = sin-1  a;;  then 


v = — % (x2  + 2)  V 1 — x2  (by  Ex.  1),  etc. 


x 3 


y = — i (a?  + 2)  a/  1 — x2  • sin-1  a;  + - + \x. 

j xtdx  . , 

dy  — — sin-1  a;. 
a/1  - a2 

xi 


[ — l{xz  + %x)  Vi — a2  + fa  sin-1  a:]  sin'1  a;  + Jg  + A3'’3* 


318 


EXAMPLES. 


53.  dy  — 


dx 

2 + cos  x 

2 


(Art.  162.) 


y — — - tan-1 
V3 


l x' 

LVI  ° l 


54.  dy  - 

y = 


x* 2dx 


V2  — 2x  + x2 
x + 3 


(See  Formula  43,  p.  345.) 


2 


(2  — 2x  + x2)i 


+ i log  [x  — 1 + (2  — 2x  + x2)i]. 


CHAPTER  V 


INTEGRATION  BY  SERIES SUCCESSIVE  INTEGRA- 
TION   INTEGRATION  OF  FUNCTIONS  OF  TWO 

VARIABLES — AND  DEFINITE  INTEGRALS. 


163.  Integration  by  Series. — The  number  of  differ- 
ential expressions  which  can  be  integrated  in  finite  terms  is 
very  small  ; the  great  majority  of  differentials  can  be  inte- 
grated only  by  the  aid  of  infinite  series.  When  a differen- 
tial can  be  developed  into  an  infinite  series,  each  term  may 
be  integrated  separately.  If  the  result  is  a converging 
series,  the  value  of  the  integral  may  be  found  with  sufficient 
accuracy  for  practical  purposes  by  summing  a finite  number 
of  terms ; and  sometimes  the  law  of  die  series  is  such  that 
its  exact  value  can  be  found,  even  though  the  series  is  infi- 
nite. This  method  is  not  only  a last  resort  when  the 
methods  of  exact  integration  fail,  but  it  may  often  be  em 
ployed  with  advantage  when  an  exact  integration  would 
lead  to  a function  of  complicated  form ; and  the  two  methods 
may  be  used  together  to  discover  the  form  of  the  developed 
integral. 

EXAMPLES. 


1.  Integrate 


, dx 

dy  = m a series. 

a + x 


By  division, 


I 1 x ct?  x? 

i ~ T ~ a kT  GtCi 

a + x a a 2 a3  a4 


/*  dx 
J a + x 


x x*  Xs  \ , 

+ . + etc. ) dx 

a 2 a3  a4  / 


320 


EXAMPLES. 


But 


f, 


dx 

a + r 


X X*  Xs  05* 

~~  a ~ 2cd  + 3a*  ~ 4^  + etc‘ 


= log  ( a + x).  [Art.  130,  (4).] 


x xi 


7? 


x 4 


logO  + ^)  = --^  + ^-£i  + et0. 

2.  dy  = x-  (1  — .r2)i  dx. 

Expanding  (1  — x2)*  by  the  Binomial  Theorem,  we  have 

, /y*2  ryA  /y»6  /y*8 

(1  _ _ _ Pte 

[ ~ 2 8 16  128  1 


••  y = fxi(1~l 


x2  xi  a:6  ox8 


etc 


.J  dx. 


2 8 16  128 

= \xl  — \x*  — i \x  ~ — rhjx  ~ — xi t$x  — e^c< 
dx 


*•  dy  = j 

y = fi 


+ x2 

dx 
-f-  x2 


= tan-1  x [Art.  131,  (16)] 


Xs  X5  X1  X 9 

- * — 3 + 5 ~ 7 + 9 “ etC‘ 


A a dx 

1 dy  = , • 

Vl  + x2 


= / 


dx 


Vl  + x1 


= log  ( X + Vl  + **) 


rz3 


3X5 


3-5z7 


~X  2-3  + 2-4-5  2-  4-  6-  7 


[Art.  144,  (1)] 
+ etc. 


5.  dy  — x ^ (x  — 1)5  dx. 

y = — 4V  + — etc. 


SUCCESSIVE  INTEGRATION. 


321 


164.  Successive  Integration.  — By  applying  the  rules 
previously  demonstrated  for  integration,  we  may  obtain 
the  original  function  from  which  second,  third  or  nih  dif- 
ferentials, containing  a single  variable,  may  have  been 
derived. 


cfiy 

If  the  second  derivative  = X be  given,  when  X is 


any  function  of  x,  two  successive  integrations  will  be 
required  to  determine  the  original  function  y in  terms  of  x. 
Thus,  multiplying  by  dx,  we  have 


2 = Xdx •• 


Integrating,  we  get 

| = fm  = x,  + «. 

Multiplying  again  by  dx  and  integrating,  we  get 
y = J ' Xxdx  -f  J*  Cidx  = X2  + Cxx  + Cv 


cfiy 

Similarly,  if  we  had  — X,  three  successive  integral 
tions  would  give 


y = X3  + C}^r  + C?x  + C3,  and  so  on. 


Generally,  let  there  be  the  nth  derivative 


322 


SUCCESSIVE  INTEGRATION. 


hence,  by  integrating  we  have 

3SF?  = fxdx  = * + c- 

Again,  we  get  from  this  last  equation, 

= Xxdx  + C\dx  ; 

= + Cxx  -f  Cv 

obtain 

= X^dx  + C\x  dx  -J-  C,dx, 

= X3  -f-  Clg"  + Cj X + 

And  continuing  the  process  we  get,  after  n integrations, 

n /»n 

dny  — J Xdxf1 

= Xn  + Ci  1-2-3  . . . (n  - 1)  + °*  1-2-3  ...  (n  — 2) 

+ - - - - Cn_,x  + C..  (1) 

The  symbol  f Xdxn  is  called  the  nth  integral  of  Xdxn, 

and  denotes  that  n successive  integrations  are  required. 
The  first  term  Xn  of  the  second  member  is  the  nfl1  integral 
of  Xdxn,  without  the  arbitrary  constants  ; the  remaining 
part  of  the  series  is  the  result  of  introducing  at  each 
integration,  an  arbitrary  constant. 


and  by  integrating, 

dn~2y 
dx n~z 

Also  from  this  we 


and  integrating, 

dn~3y 

dxn~3 


DEVELOPMENT  OF  INTEGRALS. 


323 


/‘n 

165.  To  Develop  the  nth  Integral  f Xdxn  into 
a Series. — By  Maclaurm’s  theorem,  we  have 

fHXdxn  = {J'n  Xdx^  + (f^Xd^fl 

+ (/  + 


-4- 


+ 


+ 


(A* *)  r 

ldX\  'X 

\ dx  ) 1 • 2 . . . 

m\ 


2-3  ...  (re  — 1) 


xn 


1-2-3  ...  n 


>n-|- 1 


(n  + 1) 


\ dx 2 / 1 • 2 . . . (re  + 2) 
in  which  the  brackets 


+ etc. 


(1) 


( fnXdx^ , (/  Xd*-1)  ....  (f  Xdx), 


are  the  arbitrary  constants 

Cn,  CL„ 


- a, 


for  that  is  what  these  expressions  become  respectively, 
when  x = 0. 

By  Maclaurin’s  theorem,  we  have 


X-.~ 


. (dX\x  id2X  \ x2  (d3X\  x3  . 

+ b)  I + br)  2 + (*?)  273  + <"*•  (3) 


which  may  be  converted  into  (1)  by  substituting  for 
x°,  x1,  x2,  x3,  etc.,  in  (2),  the  quantities 

/y»W  /yiTJ-h  2 

^rr,  etC., 


1-2...  re’  2-3...  (re + 1)’  3-4...  (re  + 2) 


d /*“  /*"— 1 

* Since  — / Xto"  = / Xc/a:'*'1. 


324 


EXAMPLES . 


and  prefixing  tne  terms  containing  the  arbitrary  constants 
as  above  shown,  viz., 

rp  /p  2 rgOfl — 1 

n n n * p ^ 

G„,  tMl,  ....  1.2. 3 — 1)' 

(See  Lacroix,  Calcul  Integral,  Yol.  II,  pp.  154  and  155.) 


EXAMPLES. 


1.  Develop  J 


4 dxi 


Vl  — x2 
Here  X = (1  — x2)~i 

= 1 + ix?  + 271**  + ^6^  + etc- 

Substituting  in  this  series  for  ad,  x2,  ad,  ad,  etc.,  tbs 
quantities 

x4  x 6 ad  a.10 

etc  * 

1- 2-3-4’  3- 4- 5- 6’  5-6-7-8’  7-8-9-10’  ’ 

and  prefixing 


p p ^ n x~  /-r  x~ 

°4’  °31’  tal-2’  1-  2-  3* 

we  get 

Z14  ^ Z7  i r ^ i C x*  i C 

J 4+  3i+  ii72+  '1-2- 

ad 


a;4 


a;3 


1- 2-3- 4 + 2- 3- 4-5-6  ^ 2- 4- 5- 6- 7- 8 
l-3-5ad° 


ad 


+ 


3 

1-  3ad 


+ 


+ etc. 


2-  4-  6- 7-8- 9- 10 
2.  Integrate  d3y  = 6 a cZad. 
Dividing  by  dx 2 we  have 

%=udx’  °r  <*(§0 


= 6u  tZa;. 


EXAMPLES. 


3 Zb 


or 


d2y 

dx2 


— Qax  + C,. 


Multiplying  by  dx  and  integrating  again,  we  have, 


dy  _ 


dx 


— 3 ax2  + Cxx  + Ct. 


Multiplying  again  by  dx  and  integrating,  we  have 
y = ax3  -f-  Ci  — + C.2x  -f-  Cg. 

3.  Integrate  d2y  = sin  x cos2  x dx2. 

Dividing  by  dx,  we  have 

ddy 
dx 

Integrating,  wre  have 


= sin  x cos 2xdx. 


dy 


— — cos3a;-f-  Cj 


dx~  3 ~ 1 1# 

Multiplying  by  dx  and  integrating  again,  we  have 

sin3  a;  sin  a:  ^ ^ 

V = —9 3-  + Gtx  + C2 

4.  Integrate  d3y  = ax2dx3. 

ax5  Cxx2  ri 

y ~ 60  + T~  + CiX  + Ci' 


5.  Integrate  d3y  = 2 x~3dx3. 


y = log  ^ + -g-  + ^ + CP 
d*y  — cos  x dxt. 

Cyx2 

y = cos  x + — - + + Cga;  + C, 


6.  Integrate 


326 


INTEGRATION  OF  FUNCTIONS. 


166.  Integration  of  Functions  of  Two  or  More 
Variables. — Differential  functions  of  two  or  more  vari- 
ables are  either  partial  or  total  (Art.  80).  When  partial, 
they  are  obtained  from  the  original  function  either  by 
differentiating  with  respect  to  one  variable  only,  or  by 
differentiating  first  with  respect  to  one  variable,  regarding 
the  others  as  constant;  then  the  result  differentiated  with 
respect  to  a second  variable,  regarding  the  rest  as  constant, 
and  so  on  (Art.  83).  For  example, 


and 


t Pu 


dx  dy 


f(x>  y) 


are  differential  functions  of  the  first  and  second  kinds 
respectively,  in  which  u is  a function  of  the  independent 
variables  x and  y.  From  the  manner  in  which  the 
expression  cPu 

dx 3 


= /(«»  y ) 


was  obtained  (Art.  83),  it  is  evident  that  the  value  of  u 
may  be  found  by  integrating  twice  with  respect  to  x,  as  in 
Art.  165,  regarding  y as  constant;  care  being  taken,  at 
each  integration,  to  add  an  arbitrary  function  of  y,  instead 
of  a constant. 

167.  Integration  of  dy  ^ = f(x,  y). 

This  equation  may  be  written 

It  is  evident  that  -r-  must  be  a function  such  that  if  we 
dx 

differentiate  it  with  respect  to  y,  regarding  x as  constant, 
the  result  will  be  / (x,  y). 

Therefore  we  may  write 

% = . /}>>  y)  dy- 


INTEGRATION  OF  PARTIAL  DIFFERENTIALS.  327 


Here,  also,  it  is  evident  that  u must  be  such  a function 
that  if  we  differentiate  it  with  respect  to  x,  regarding  y as 
constant,  the  result  will  be  the  function 


Therefore,  we  first  integrate  with  respect  to  y,  regarding 
x as  constant,*  and  then  integrate  the  result  with  respect  to 
x,  regarding  y as  constant,*  which  is  exactly  reversing  the 
process  of  differentiation.  (Art.  83.) 

The  above  expression  for  u may  be  abbreviated  into 

J ff  (-A  V)  dy  dx  °r  J f f (A  y)  dx  dy. 

We  shall  use  the  latter  form;f  that  is,  when  we  perform 
the  y-integration  before  the  ^-integration,  wre  shall  write  dy 
to  the  right  of  dx. 

It  is  immaterial  whether  we  first  integrate  with  respect  to 
y and  then  with  respect  to  x,  or  first  with  respect  to  x and 
then  with  respect  to  y.  (See  Art.  84.) 

In  integrating  with  respect  to  y,  care  must  be  taken  to 
add  an  arbitrary  function  of  x,  and  in  integrating  with 
respect  to  x to  add  an  arbitrary  function  of  y. 

In  a similar  manner,  it  may  be  shown  that  to  find  the 
value  of  u in  the  equation 


Hence, 


dx  dy  dz  ~ f(X’  y’ 


we  may  write  it 


* Called  the  ^-integration  and  2-integration,  respectively, 
t On  this  point  of  notation  writers  are  not  quite  uniform.  See Tofitimter’s  Cd  , , 


328 


EXAMPLES. 


which,  means  that  we  first  integrate  with  respect  to  z,  regard- 
ing x and  y as  constant;  then  this  result  with  respect  to  y , 
regarding  x and  z as  constant ; then  this  last  result  with 
respect  to  x,  regarding  y and  z as  constant,  adding  with  the 
2-integration  arbitrary  functions  of  x and  y,  with  the 
^-integration  arbitrary  functions  of  x and  z,  and  with  the 
^-integration  arbitrary  functions  of  y and  z.  (See  Lacroix, 
Calcul  Integral,  Vol.  II,  p.  206.) 


EXAMPLES. 

1.  Integrate  d?u  — bx2ydx2 


Here 


du 

• 

dx 


2.  Integrate 


= J'bxhydx  = \bx?y  + f{y). 

du  = \bx?ydx  -f  / (y)  dx. 
u — ^btfy  + f(y)  x + <p(y). 
d2u  = 2x2y  dx  dy. 


Here  d — 2x2ydy. 

•••  ^ = fWydy  = ~>hf  + (p  (x). 

du  = x2y2dx  + (f)  (x)  dx. 
u — p3y2  + f(p  {■>■)  dx  +f{y). 

3.  Integrate  dhi  = 3 xyz  dx  dy. 

u — PY  + f<t>  (x)  dx  +f{y). 

4.  Integrate  (Pu  = ax3y2  dx  dy. 

a 


— p2  ^yz  + (x)  dx  + f(y). 


INTEGRATION  OF  TOTAL  DIFFERENTIALS.  329 


168.  Integration  of  Total  Differentials  of  the  First 
Order. 


If 

we  have  (Art.  81), 


* = f(x>  y). 


7 du  , du  , 
du  = -^dx  dy, 

du  du 

in  which  ^ dx  and  dy  are  the  partial  differentials  of  u ; 
also,  we  have  (Art.  84), 


(Pu 


cPu 

dy  dx’ 


or 


dx  dy 

d / du\  d / du\ 

dy  \dx)  dx  \dy) 


Therefore,  if  an  expression  of  the  form 
du  — Pdx  + Qdy 

be  a total  differential  of  u,  we  must  have 


(1) 

(2) 


du  _ „ 

di~  1 ’ 


du  _ _ 

dy~ 


(3) 


and  hence,  from  (1),  we  must  have  the  condition 

dP  _ dQ 
dy  ~~  dx  ’ 

which  is  called  Eider’s  Criterion  of  Integr ability.  When 
this  is  satisfied,  (2)  is  the  differential  of  a function  of  x and 
y,  and  we  shall  obtain  the  function  itself  by  integrating 
either  term  ; thus, 

u = /Pdx+f(y),  (4) 

in  which  / (y)  must  be  determined  so  as  to  satisfy  the  con- 
dition 

du 

dy~ 


330 


EXAMPLES. 


Remark. — Since  the  differential  with  respect  to  x of  every  term  ol 
u which  involves  x must  contain  dx,  therefore  the  integral  of  Pdx  will 
give  all  the  terms  of  u which  involve  x.  The  differential  with  respect 
to  y of  those  terms  of  u which  involve  y and  not  x,  will  be  found  only 
in  the  expression  Qdy.  Hence,  if  we  integrate  those  terms  of  Qdy 
which  do  not  involve  x,  we  shall  have  the  terms  of  u which  involve  y 
only.  This  will  he  the  value  of  f(y),  which  added  with  an  arbitrary 
constant  to  J Pdx  will  give  the  entire  integral.  Of  course,  if  every 
term  of  the  given  differential  contain  x or  dx,  f(y)  will  he  constant. 
(See  Church’s  Calculus,  p.  274.) 


EXAMPLES. 

1.  du  = fc$y3dx  -f  3z4y2dy. 

Here  P — 4a3y3,  Q — 3xiy2. 

:.  — 12  x3y2  and  = 12  x?y2. 

dy  J dx  ^ 

Therefore  (3)  is  satisfied,  and  since  each  term  contains  2 
or  dx,  we  have  from  (4), 

u — f 4:X3y3dx  = xiy3  + C. 

2.  du  = j+  (2y-X-^dy. 

(3)  is  satisfied,  therefore  from  (4)  we  have 

U = fdy+f{y)  = 

Since  the  term  2 ydy  does  not  contain  x,  we  must  have, 
from  the  above  Remark,  / (y)  — f2ydy  = y2,  which  must 

X 

be  added  to  giving  for  the  entire  integral, 

x 2 n 

u = - + '/+ C- 

3.  du  — ydx  4-  xdy.  u = xy  + C. 

4.  da  — (6 xy  — y2)  dx  + (3a;2  — 2 xy)  dy. 

u — 3 a?y  — y2x  + C. 


DEFINITE  INTEGRALS. 


331 


5.  du  = (flaxy  — 3bx2y)  dx  + (ax2  — la?)  dy. 

u = ax2y  — byx?  + C. 

The  limits  of  this  work  preclude  us  from  going  further  in 
this  most  interesting  branch  of  the  Calculus.  The  student 
who  wishes  to  pursue  the  subject  further  is  referred  to 
Gregory’s  Examples;  Price’s  Calculus,  Yol.  II;  Lacroix’s 
Calcul  Integral,  Yol.  II ; and  Boole’s  Differential  Equations, 
where  the  subject  is  specially  investigated. 


169.  Definite  Integrals.—  It  was  shown  in  Art.  130 

that,  to  complete  each  integral,  an  arbitrary  constant  C 
must  be  added.  While  the  value  of  this  constant  C remains 
unknown,  the  integral  expression  is  called  an  indefinite  in- 
tegral ; such  are  all  the  integrals  that  have  been  found  by 
the  methods  hitherto  explained. 

When  two  different  values  of  the  variable  have  been  sub- 
stituted in  the  indefinite  integral,  and  the  difference  between 
the  two  results  is  taken,  the  integral  is  said  to  be  taken 
between  limits. 

In  the  application  of  the  Calculus  to  the  solution  of  real 
problems,  the  nature  of  the  question  will  always  require 
that  the  integral  be  taken  between  given  limits.  When  an 
integral  is  taken  between  limits,  it  is  called  a definite 
integral.* 

The  symbol  for  a definite  integral  is 


which  means  that  the  expression  f(x)  dx  is  first  to  be  inte- 
grated ; then  in  this  result  l and  a are  to  be  substituted 
successively  for  x,  and  the  latter  result  is  to  be  subtracted 
from  the  former  ; l and  a are  called  the  limits  of  integra- 
tion, the  former  being  the  superior,  and  the  latter  the 
inferior  limit.  Whatever  may  be  the  value  of  the  integral 

* In  the  Integral  Calculus,  it  is  often  the  most  difficult  part  of  the  work  to  pass 
from  the  indefinite  to  the  definite  integral. 


332 


DEFINITE  INTEGRALS. 


at  the  inferior  limit,  that  value  is  included  in  the  value  ol 
the  integral  up  to  the  superior  limit.  Hence,  to  find  the 
integral  between  the  limits,  take  the  difference  between  the 
values  of  the  integral  at  the  limits. 

In  the  preceding  we  assume  that  the  function  is  continu- 
ous between  the  limits  a and  b,  i.  e.,  that  it  does  not  become 
imaginary  or  infinite  for  any  value  of  x between  a and  b. 

Suppose  u to  be  a function  of  x represented  by  the  equa- 
tion 

u = f{x)\ 

then  du  — f ( x ) dx. 

Now  if  we  wish  the  integral  between  the  limits  a and  b , 
we  have 

u=  f /'(ar)«fo  =/(&)— /(a). 

If  there  is  anything  in  the  nature  of  the  problem  under 
Consideration  from  which  we  can  know  the  value  of  the 
integral  for  a particular  value  of  the  variable,  the  constant 
C can  be  found  by  substituting  this  value  in  the  indefinite 
integral.  Thus,  if  we  have 

du  = {abx  — bx2)  -•  (ab  — 2 bx)  dx, 

and  know  that  the  integral  must  reduce  to  m when  x = a, 
we  can  find  the  definite  integral  as  follows: 

Integrating  by  known  rules,  we  have 

u — f {abx  — bx 2)v  q-  C, 

which  is  the  indefinite  integral ; and  since  u —m  when 
x = a,  we  have 

m — 0 + (7;  .-.  C — m, 

which  substituted  in  the  value. of  u gives 
u ~ \ ( abx  — bx 2)^  + m. 


EXAMPLES. 


333 


EXAMPLES. 


1.  Find  the  definite  integral  of  du  = (1  + | ax^  dx,  on 
the  hypothesis  that  u — 0 when  x = 0. 

The  indefinite  integral  is 


U ~ 27a  ^ + & 


Since  when  x — 0,  u = 0,  we  have 


0~2^+C’ 


0=  - 


which  substituted  in  the  indefinite  integral,  gives 

8 / . s 8 

U = 27^  (1  + ^ ~ Wa 

br  the  definite  integral  required. 

2.  Integrate  du  = 6x~dx  between  the  limits  3 and  0. 


Here 


u = C 6x2dx  = 

J o 

n\  r-  2^+1  - 

3.  u — xndx  = — —7 

t/o  (_«  4-  1_ 

4.  u — J‘  e~x  dx  = — e_I 


2a^ 

l 

o 


3 * 


= 54. 


n + 1* 

= -(0-l)  = L 


5 . u = f 

r>a 

6.  u = / - 

«'o  «' 


00  rf* 


o a2  + Xs 
dx 


= t r tao-1  -~| 
a « 


tan  1 ' 


7T 

2 a 

7 r 

4a* 


* This  notation  signifies  that  the  integral  is  to  be  taken  between  the  limits  3 
and  0. 


334 


CHANGE  OF  LIMITS. 


7.  U 


dx 

a 2 -f-  x1 


1 

a 


tan-1 


x 

a 


= - [tan-1  co  — tan-1  (—  co  )1  = - 
a L v /J  a 


ra  dx 

8.  u — / — - — - - 
^ o V«2  — as* 


a 

0 


Remake. — It  should  be  observed  here  that  the  value  of  the  infini- 
tesimal element  corresponding  to  the  superior  limit  is  excluded,  while 
that  corresponding  to  the  inferior  limit  is  included  in  the  definite  in- 
tegral ; for,  were  this  not  the  case,  as  — — ^ becomes  equal  to  go 

y«2  - a* 

when  x — a,  the  integral  of  Ex.  8 between  the  limits  a and  0 would 
not  be  correct ; but  as  the  limit  a,  being  the  superior  limit  in  Es.  8, 
and  that  which  renders  infinite  the  infinitesimal  element,  is  not 
included,  the  definite  integral  is  correct.  (See  Price’s  Calculus, 
Vol.  IT,  p.  89.) 


9.  u = f (a2  — x2)i  dx.  (See  Ex.  4,  Art.  151.1 


a1 7r 


j<«! -**)»  + 2 ,o=  4 


10.  u 


/*i  cfidx 

= / —===■  (See  Ex.  7,  Art.  151.) 

do  W 1 _ a;2  v 


Vl  — 

1.3-5-tt 
“ 2-4- 6-2° 


11.  u 


= /.  i 

Co 


sin7  x cos4  x dx  = 


42 

3-5-7-11' 


170.  Change  of  Limits. — It  is  not  necessary  that  the 
increment  dx  should  be  regarded  as  positive,  for  we  may 
consider  x as  decreasing  by  infinitesimal  elements,  as  well  as 
increasing.  Therefore,  we  have 


CHANGE  OF  LIMITS. 


335 


/ < p ' (, x ) dx  — 4>  (a)  — 0 (b)  = — [<t>  ( b ) — («)] 
d b 

— — f <p'  (x)  dx. 

v a 

That  is,  if  we  interchange  the  limits,  we  change  the 
sign  of  the  definite  integral. 

Also,  it  is  obvious  from  tlie  nature  of  integration  (Art.  129),  that 

pc  pb  pc 

/ <j>(x)dx  = / <p(x)dx  + / <p  (x)  dx, 

da  da  d b 


and  so  on.  Hence, 


p n pi n pin  p\n 

/ cos  x dx  = / cos  x dx  + / cos  x dx  + / cos  x dx 

do  do  d in  ^ \n 

+ / cos  x dx 
din 

r nn 

= / cos  x dx  + / cos  x dx  = 0. 

* J 0 */ 

/>0  /»<z 

(x)  d*. 


(1) 


/a  /»0 

/'  (®)  dx=  f (x)  dx+  f 
a —a  0 

Let  x — — x ; then  dx  — — dx,  and  the  limits  0 and  — a become  0 
and  + a ; therefore  we  have 

p0  ,i0  pa 

/ f'(x)dx=—  / f'{-x)dx-  / f'(-x)dx, 

d —a  da  do 

which  in  (1)  gives 

/ia  pa  pa 

f'(x)dx-  / f'(~x)dx+  / /'(*)<&; 

-a  i/  0 t ' 0 

= ^ [/'  (—as)  dx  + /'  (®)  (f®J. 

Now  if  f (—x)  — — f (x),  (2)  becomes, 

/a 

f ( x ) dx  = 0. 


(2) 


336 


EXAMPLES. 


But  if  /'  (— x ) —f  ( x ),  we  liave 

/O  pa 

f (*)  dx  — / /'  ( x ) dx, 

r*a  r*a 

which  in  (1)  gives  / /'  (x)  dx  — 2 / f (x)  dx. 

The  following  are  examples  of  these  principles. 

->$7r 


(3) 

(4) 


1.  u 

Here 


cos  x dx. 

Jrr 

/'  ( — x)  = f (x)  — cos  X. 

/(Jtt  rt  in- 

COS  x dx  — 2 / cos  x dx  — 2. 

•i>r 

2.  u = I sin  x dx  — 0.  [Since/' ( — a;)  == — /'(a;).] 

3.  u = S'*  (a*  — a?)*  dx  = 2 f*  (a?  - x*)±  dx  = 

/J?r  *-7T  /'•TT 

u=  sin  x dx  = / sin  x dx  -f-  / sin  x c?x 
<d  0 ^0  «'hr 

— 21  sm  x ax. 

t/n 


jjSince  sin  x dx  = j*  si 


sin  x rfx 


- 2. 


/*7T  />^7T  /*7T 

5.  u=  cos  x dx  — / cos  x <?x  + / cos  x c?x  = 0. 

o 'dip 

Since  / cos  x dx  — — / cos  x <7x. 

i/rr  ^0 


6.  m 


arWx  _ 2 xidx 

J—l  \/l  — XJ  1 0 \/l  — X2 


= 


(See  Ex.  2_.  Art.  162.) 


EXAMPLES. 


33? 


EXAMPLES. 


1.  Integrate  du  — — 


<?a; 


V«2  — x2 


by  series. 


x 

u = - + 


a; 


3 1 • 3a:5  1 • 3 • 5a:7 

+ o „ c„«  + + etc. 


a ' 1-2-3 a3  1 2-4-5a5  1 2-4-6-7a7 

But  f — —X = sin-1-  (by  Ex.  14  of  Art.  131); 

J Va2  — a:2  a 

therefore, 

. x x a3  l-3a;5  1-3 -5x7 

Sm  a ~ a + A3V  + 2T4T5H*  + 2^4- 6- 7a7  + etC' 

2.  Integrate  rfw  = — - .. 

0 Vi  + z4 


1 . 3 . 5^13 

“ = I ~ 2^5“  + 2TE9  _ 2-4-6-13  + etC" 


x 1 • x5  . 1 • 3a:9 

nr 

dx 


3.  Integrate  — 

V 1 — -a:2 

By  the  Binomial  Theorem, 


(1  — *?)*. 


1 129  l-e4^  l-3e6a:6  , 

VI  - ^a;2  = 1 - ^e2x2  - - - etc. 


Multiplying  by 


dx 


Vl  — x2 

arately  (see  Ex.  1,  Art.  151),  we  have 


, and  integrating  each  term  sep- 


r dx 

u = J 


\/l  — x2 


(1  — e2x2)i 


= sin  1 x + \e2 


2 Vl  — x2  — \ sin 


-7a:J 


e4  P /.r3  1 • 3 • a:\  1-3  . , 1 

+ 2:4  Lt + Vl  - ^ - 2 7i sm  27 J 


15 


338 


EXAMPLES. 


, 1 - 3 • e6 
+ 2-4-6 


/xr‘  l-5x3  l-3-5x\  r 

_(e  + T-T  +2^6  j 

1-3-5 


1- o-o  . | 

2- 4- 6 Sm  XJ 

2 

4.  Given  cPy  = — > dr?,  to  find  y.  (Art.  164.) 


+ etc. 


y = logo;  + |G,a;2  + G^  + G3. 

5.  d4y  = — ar4rfar*. 

V = i log  a:  + £G,a?  + |G2a;2  + G3a;  + G4. 

6.  = x^dx?.  y — xot-^  ^ 

7.  (Py  = Sx2dx2.  y = -feSz4  + C\x  -f-  Gj. 

8.  d2y  = cos  x sin2  x dx2. 

y = \ cos3 a:  — -^-cosa;  + C.,  x + C2. 

9.  d*y  — cos  x dx4. 

y = cos  a:  + ^C\'jP  -f  \C&2  -f-  C3x  -f  G4. 

10.  dsy  = e~dxz.  y = ex  + \CxxP  -j-  C/c  -f-  G3 

12  d4y  — (1  + x2)~^  dx4.  (Art.  165.) 

^c.  + az  + c^  + c^  + AL 

x£  , l-3a^ 

2-3-4- 5- 6 + 2-4-5- 6^7-8 


1 • 3 • 5.r10 

2-4- 6- 7-8- 9-10  + GtC" 

12.  cPu  = axPtfdxdy. 

u = 30  ^ (x)  dx  +/(y)- 


13.  du  — (2 xy2  + 9 x2y  + 8x3)  dx  + (2 xly  4-  3a^)  dy. 
u = xhy2  + 3 x?y  + 2ar*  + G. 


EXAMPLES. 


339 


14.  u = / (a2  — a:2)  2 (fa  = 2 / (a2  — x2)%  dx. 

v —a  P o 

(See  Art.  170.) 

u — a6^.  (See  Ex.  7 of  Art.  162.) 


xb. 


» = / 


2a  xidx 


v — 


-v/  2 ax  — x* 

7-5-3 
4-3-2 


crn. 


16.  » = *)* ix  = 3,,.^- 

17.  « ^ P ~J*L=  (1  - eV)i  (See  Ex.  3.) 

t/o  -\/ 1 — a;2 


7r  1 „ rr  1-3  . n 1-32-  5e6  n 


tt  = r--e‘1-- 


4 


18.  u = 


2 4 2 22-42  2 22-  42-  62  2 

/iiir  s>x  x dx  dy 


— etc. 


, , 2 , {•  (Art.  167.) 

'o  %2  + V2 


We  first  perform  the  ^/-integration,  regarding  x as  con- 
stant, and  then  the  a*-integration. 


u = 


- tan-1  y- 
x x 


x dx.  (Art.  132,  Ex.  3.) 


19.  u = 


Pi*  TT  7T2 

= J„  4^=16 

xyz  dx  dy  dz 


rr  rj2 

'o  * 16 

,->a  nx  ny 
’ 0 ’’  0 d Q 

1 a x xid  , , Pa  X?  , a6 

o .0  T*«  = J.  8 dx  = ls 


pa  pa-x  pa—x-y  #3 

20.  u = / / / dx  dy  dz  = — « 

«^o  o'o  v> 


340 


FORMULA  OF  INTEGRATION. 


Ft  CL  /Tr4  j 

21.  u — f / r3dd  dr  — 
a " n 


22.  u 


o "o 

\n  nia  co8 

n<I  0 


2 


= fJ 

1/  _ 1 jrU  n 


r3dd  dr  — laKn. 


For  the  convenience  of  the  student,  the  preceding  for* 
mulse  are  summed  up  in  the  following  table. 


TABLE  OF  INTEGRALS. 

CHAPTER  I. 

Elementary  Forms.  (Page  23 SO 

1.  J 1 ( dv  + dy  — dz)  = v + y — z. 


S.  /< 


axTdx  — 


ax' 


.n+1 


T 1 

, Padx  _ a 

J of-  (n  — 


(7i  — 1) 

. Padx  . 

4 J — = alosx- 

5.  J* ax  log  adx  — cf. 

6.  f exdx  = e*. 

7.  J* cos  x dx  = sin  x. 

8.  f sec2  x dx  = tan  x. 

9.  f sec  x tan  x dx  = sec  x. 


(130)* 

(131) 


arbitrary  constant  is  understood.  (See  Art.  131.) 


FORMULA  OF  INTEGRATION. 


341 


10.  / 


dx  1 . bx 

= 7 sin  1 — 

V a,1  — b2x2  b a 


ii  r dx  1 bx 

1L  J a > + Vx‘  - a i ~a’ 


18.  / 


dx  1 . bx 

— S6C  * • 

x s/b2B  — a2  a a 


13 


r d 

a / 0 n h i' 


dx 


1 , bx 

z:  = 7 vers-1  — • 

V 2abx  — b2x2  ® a 


U-  f tan  x dx  — log  sec  x. 

15.  / ^X—  = log  tan  \x. 

J sm  x ° * 


(132) 


16. 


17. 


18. 


CHAPTER  II. 

Rational  Fractions.  (Page  256.) 

(x)  dx  f‘  Adx  Bdx  Ldx  . 

J- jk  = J T-a  + ~i+  ■ ■ J —x<™) 

Pf(x)  dx 

J <p  (x) 

p Adx  p Bdx  p Ldx  . 

J (x — ay,Jr J (x — (x — a)  ' 


(; x—af  J (x — a)n~l  ' J (x — a) 

pf(x ) dx  _ p (Ax  -h  B ) dx 

J <p  (x)  J [(a;  + a)2  + Wp 

n— 1 "P 


p (Cx  + D ) dx  p(Ex  + L ) dx 

* J {{x  ± af  + Z*2]”-1  + ' ' ‘ J (x  ±af  + b 2 


(139) 


*>■ 


ix  — a 


x + a 


342 


FORMULAE  OF  INTEGRATION. 


23. 


24, 


CHAPTER  III. 
Irrational  Functions.  (Page  269.) 
p xndx  2 P . „ , 

d Va  + bx  bn+lJ  K ' 

[where  z = [a  + bx)?]. 

P x2n+ldx  _ P (z2  — a)n  dz 
' [a  + bx2)^  ” ^"+1 

[where  z — (a  + bx2)^\ 
dx 


21 


(142) 


\/ a bx  -\-  x2 


(143) 

= log^  + rc+A/a  + to  + a;2)-  (144) 


dx 


— 2 tan-1 


(3  — a; 


or 


f 

, f- 

^ V a + bx  — x 2 V x — a 

[where  a + bx  — x2  = (x  — «)  (0  — x ). 

25.  J'i xm(a  + bxPydx  = ^ (146> 

(where  2?  — a + 5a;”) ; 

2a(-+^)  f (#-b)^+I*+')  dz, 

n v 

(where  xnP  = a + bxn). 


26 


dx 


2 / Va-p&r— V«\ 

Va  °g  V * /' 


. f-  

d x's/a  + bx 
27.  f . -S^_===llog  (to+Vg2  + #^2)- 
p dx  1 , (V< n2  + &2r2 — <A 


Ex.  6,  (146) 
Ex.  16 


TV  17 


FORMULAE  OF  INTEGRATION. 


343 


29.  f — z=~===  — a log  (x-\-a  + V%ax  + x2).  Ex.  20. 

J V2az  + x 2 

30.  J (a2  + 22)i  dx 

=■  | («2  + z2)*  + 1-  log  [x  + (a2  + z2)^].  Ex.  35. 


CHAPTER  1 Ar. 

Successive  Reduction.  (Page  285.) 

31.  J udv  — uv  — J vdu.  (147) 

32.  J xm  (a  + bxn)p  dx  (148) 

xm-n+i(a-\-b7f')p+ 1 — (m — n + 1)  a f x"l~n(a  + bxn)pdx 


b ( np  + m -f-  1) 
33.  j * x~m  ( a + bxnY  dx 


(-1) 

(149) 


x~m+\a  + bxny-1  + b(m  — np  — n — 1)  i x~m+n (a  4-  bxn)vdx 

: — ^ .(B) 

— a (in  — 1) 

34.  J* xm  ( a + bxn)p  dx  (150) 

zm+1  ( a + bzn)p  + anp  j xm  (a  + 5z")y_1  dx 


np  m \ 

35.  j xm  {a  + bxn)~P  dx 
xm+l(a  + bx,l)~p+l  — (m  + n + l—np)  f xm(ci  + bxn)~^+ldx 


(C) 

(151) 


36 


an  (p  — 1 ) 
. t f (ft2  — x2)~i  dx 

(in  — 1) 


{D) 


= — («2-z2)i  + «2  f xm~2  («2  ~ dx. 


in 


344 

37. 

38. 


39. 


FORMULA  OF  INTEGRATION. 

J‘ xm  (a2  + x2)~i  dx 

= — («2  + X*)i-  i™-1)  a2  f ^ (fl2  + 22)4 

p 

v xm  ( a 2 — x2)i 

— ( a 2 — x2)i  m — 2 P dx 
(m  — 1 )a2xm~'  ( m — 1 ) a2  J xm~2(a2 z2)? 

J (a2  — x2)-  dx 

x (a2  — x2)2  + na2  C ( a 2 — x2)'2~'dx 


n + 1 


40. 


/ 


(a2  — a;2)! 
x 


w— 3 p dx 

(n — 2)a2(a2 — + {n—2)a2J  (a2—x2)i~1' 


41. 


fir. 


xmdx 


-\/  2ax  — x2 


(2 m — 1)  a p xm~xdx 


42 


m 
xndx 


m 


/\/2ax  — x1 


= — V %ax  — x2  + 

/ 

= xn~l  V a + bx  + cx2  n — 1 a p xn~2dx 

co 


V a + bx  + cx2 


nc 


?i  co  a fox  c3$ 

2 n — 1 b p xn~xdx  * 
2)i  ft  \/a  + bx  + ex? 


* See  Price’s  Calculus,  Vol.  II,  p.  63. 


FORMULAE  OF  INTEGRATION. 


345 


43 


•/; 


x2dx 


V a + bx  + cx2 

= + £)  + (£~  |J/ 

. . r x dx 

44.  / - 


dx  * 


V a + bx  + cx2 


a + lx  -j-  cx1 

V a + bx  + ca;2  5 Z1  ^ 


- /- 


c 2 c " V a bx  + cx2 


45 


. J* xF  log” 


xdx 


rm+ 1 


J 


46.  / =— 


to  + 1 
’af*  r/a; 
log"  x 


log”  x — ^ - J* xm  log”-1  xdx.  (152) 


m -f  1 r xm  dx 
" 1 ” — 1 J log”-1  X 


(153) 


47, 


( n — 1)  log”-1  a;  n 

, r dz 
J log  z 

= log  (log  Z)  + log  2 + ^ log2  2 + ^32  log3 2 + etc. 


48 


•/■ 


amxxn  dx  = 


raxdx  _ ax  log  « raxdx 

J xm  (to  — l)a:m-1_^TO  — 1*/  a;'”-1  ° 

50.  J* sin™  0 cos”  0 d0 

= — f (1  — cos2  0)r  cos”  Qd  cos  0,  (156) 


amx: V 


m log  a to  log  a 


J* amxxn  1 dx.  (154) 


* See  Price’s  Calculus,  Vol.  II,  p.  63. 


346 


FORMULAE  OF  INTEGRATION. 


or 


or 


(when  m — 2r  + 1)  ; 

= J tanm  0 (1  4-  tan2  0)r~ld  tan  6, 

(when  m 4-  n — — 2 r) ; 

/n— 1 

xm  (1  — x 2)  ^ dx,  (when  x = sin  d).  (157) 


51. 


/*“ 


cos  ax  dx 


x \ ■ , \ n(n— 1)  P „ , 

= — 7r(ax  sm  ax-\-n  cos  ax) — - — - / xn~L  cos  ax  dxAlo^i] 

a 3 a2  J v ' 


_ eax  cos"  1 x (a  cos  x + n sin  x) 
a 2 4-  ft2 


52.  J e“*  cos"  x dx  — 

n (n  — 1)  (‘  „ , , 

4 h -2  / e"1  cos"-2  a:  dx. 

a 2 4-  n2  J 

53.  j sin-1  xdx  — x sin-1  x + (1  — x2)i. 
dO 


(160) 

(161) 


54. 


r. 

>J  a + b cos  6 

= Tzfa*  to-[C-rl)1]’ (,Ihen  a>h)-  (162) 


V b + a + a/ b — a tan  * 
Vi2 — «2  °^-Vb  + a — Vb — a tan 


log 


^j,  (when  a<6). 


y*TL  rp  nr-i  /-pTl~  I 

x<fc-  = a+ft_,i+cu.r^+..Qr^-1j 

. xn  IdX \ 

+ ^ 1-2-3  . . . ft  + \~dx  ) 1-2-: 


■ 3 . . . (ft  + 1) 


m\ 


\ dx 2 / 1 • 2 • 3 ...  (ft  4-  2) 


in  + etc. 


(165) 


CHAPTER  VI. 


LENGTHS  OF  CURVES. 

171.  Length  of  Plane  Curves  referred  to  Rectan- 
gular Axes. — Let  P and  Q be  two  consecutive  points  on 
the  curve  AB.  and  let  ( x , y)  be  the 


point  P ; let  s denote  the  length  of 

the  curve  AP  measured  from  a fixed 

point  A up  to  P.  Then 

F 

R 

PQ  = ds,  PR  = dx,  RQ  = dy,  ^ 

// 

Therefore,  from  the  right-angled 

triangle  PRQ  we  have 

M N 

Fig.  43. 

ds  = V dx2  + dy~ ; 


hence,  s = / V dx 2 + dy2  — f ^1  + C^j  dx. 

To  apply  this  formula  to  any  particular  curve,  we  find 
dy 

the  value  of  ^ in  terms  of  x from  the  equation  of  the 

curve,  and  then  by  integration  between  proper  limits  s 
becomes  known. 

The  process  of  finding  the  length  of  an  arc  of  a curve  is 
called  tne  rectification  of  the  curve. 

It  is  evident  that  if  y be  considered  the  independent 
variable,  we  shall  have 

s = fi1  + pA 

The  curves  whose  lengths  can  be  obtained  in  finite  terms 
are  very  limited  in  number.  We  proceed  to  consider  some 
of  the  simplest  applications : 


348 


RECTIFICATION  OF  THE  PARABOLA. 


172.  The  Parabola. — The  equation  of  the  parabola  is 

f = 2 px  ; 

dy  _p_ 

dx  y 

S = dx  = ~> 


hence, 


1 P 

or  s = -J  (p2  + y2)?dy,  (which,  by  Ex.  35  Art,  146; 

= 1yi±£+P_log(!/  + v—)  + a m 


If  we  estimate  the  arc  from  the  vertex,  then  s = 0, 
y — 0,  and  we  have 

°=|1°gi?  + c';  •••  c=—  |iog/>, 

which  in  (1)  gives 

• = + f log  (y  + 'J+*) , m 

which  is  the  length  of  the  curve  from  the  vertex  to  the 
point  which  has  any  ordinate  y.  If,  for  example,  we  wish 
to  find  the  length  of  the  curve  between  the  vertex  and  one 
extremity  of  the  latus-rectum,  y — p,  we  substitute  p for  y 
in  (2),  and  get 

5 = ipV2  +|  log  (l  + V%) 

for  the  required  length. 

We  have  here  found  the  value  of  the  constant  C by  the 
second  method  given  in  Art,  169.  We  might  have  found 
the  definite  integral  at  once  by  integrating  between  the 
limits  0 and  p,  as  explained  in  the  first  method  of  Art.  169, 
and  as  illustrated  in  the  examples  of  that  Article.  Hence, 


RECTIFICATION  OF  THE  CIRCLE. 


349 


we  need  not  take  any  notice  of  the  constant  C,  but  write 
our  result 

s — ( J ( P2  + 3/2)-  dy,  (see  Art.  169) 

and  integrate  between  these  limits. 


173.  Semi-Cubical  Parabola.* — The  equation  of  this 
curve  is  of  the  form  y 2 = ax3.  (See  Fig.  39.) 

Hence,  = § V ax  and  = \a'x. 

:.  s = J (1  -f  %ax)i  dx 

= Wa  (1  + %ax)§  + a 


If  we  wish  to  find  the  length  of  the  curve  from  A to  P, 
we  must  integrate  between  the  limits  0 and  dp  (see  Art. 
128,  Ex.  9)  ; hence, 


pip  , 

— / (1  + fa#)*  dx 

■ n 


»(1  + W! 


3 P 

0 


~Wa{1+hrap)'  ~Wa 

— [0-  + — 11=P  (3f  - 1)» 


by  substituting  - — for  a. 
Ex.  10,  Art.  128.) 


(See  Art.  125,  Ex.  1.  Compare 


174.  The  Circle. — From  x?  + y2  = r~,  we  have 

dy x 

dx  ~ y 

* This  was  the  first  curve  which  was  rectified.  The  author  was  William  Neil, 
who  was  led  to  the  discovery,  about  1660,  by  a remark  of  Wallis,  in  his  Arithmetics 
Isfinitorum.  See  Gregory’s  Examples,  p.  420. 


350 


RECTIFICATION  OF  THE  ELLIPSE. 


Hence,  for  the  length  of  a quadrant,  we  have  (since  the 
limits  are  0 and  r ), 


s 


rdx 

V r2  — x2 


which  involves  a circular  arc,  the  very  quantity  we  wish  to 
determine.  The  circle  is  therefore  not  a rectifiable  curve; 
but  the  above  integral  may  be  developed  into  a series,  and 
an  approximate  result  obtained. 

By  Ex.  1,  Art.  170,  we  have 


s 


tx  x? 
rV  + 2d3r3 


+ 


1-3.-T5  1 • 3 • 5z7 

2-4-5r5  + 2-4-6-7r? 


r 

0 


“ r{1  + 2*2 


+ 


1-3 

2.4.5 


+ 2 


3-5 


+ 


etc. ) : 


, , 1 1-3  3-5 

- 1 + 2 ^ + 2^5  + 


O.A.R.iV 


+ etc. 


By  taking  a sufficient  number  of  terms,  reducing  each  to 
a decimal,  and  adding,  we  have  tt  = 3.141592653589793  + . 
For  the  approximation  usually  employed  in  practice,  t is 
taken  as  3.1416,  and  for  still  ruder  approximations  as  3^. 


175.  The  Ellipse. — From  y 2 = (1—  e2)  ( a 2 — x2),  we  have 


!=-<> 


x xVl  — e2 

e2)-  = . _ • 

V V a2  — x2 


To  find  the  length  of  a quadrant,  we  must  integrate  be- 
tween the  limits  0 and  a ; hence. 


RECTIFICATION  OF  THE  CYCLOID. 


351 


This  integration  cannot  be  effected  in  finite  terms,  but 
may  be  obtained  by  series. 

X 

Put  - = z;  then  dx  = adz.  When  x = a,  z — 1,  and 
a 

when  x — 0,  z — 0 ; therefore  the  above  integral  becomes 
ci  . dz 


— a i (l-eV)i  — == 
v o a/1  — 22 


n (-1  12  1,3 
- a 2V  22-4a 


32.42 


l-32-5 

22-42-  62< 


— etc.  j , 


(by  Ex.  17,  Art.  170),  which  is  the  length  of  a quadrant  of 
the  ellipse  whose  semi-major  axis  is  a and  eccentricity  e. 

176.  The  Cycloid.— From  x = rvers-1-  — ^/2ry — y2, 
we  have 

dx  _ y 


dy  V 2 ry 

, 2 r 

^0 


y 


— V%r  f (2 r — y)~?  dn 

v a 


— 2 (2r)*(2r  — y)* 


JO 


= 4 r, 


which  is  ^ the  cycloidal  arc; 
hence  the  whole  arc  of  the  cy- 
cloid is  8 r or  4 times  the  diam- 
eter of  the  generating  circle. 

If  we  integrate  the  above  ex-  pig.  44. 

pression  between  y and  2 r,  we  get 

s = \/2r  / (2r  — y)~ £ dy  = 2 (2 r)^  (2 r — 

t/  v 


y )4 


= 2\/2r  (2r — y)  = arc  BP. 

But  BD  = •\/BA  x BC  = V2 r (2 r — y)  ; 

arc  BP  = 2 times  chord  BD.* 


* This  rectification  was  discovered  hy  Wren.  See  Gregory’s  Examples,  p.  421. 


352 


INVOLUTE  OF  A CIRCLE. 


177.  The  Catenary. — A catenary  is  the  curve  assumed 
by  a perfectly  flexible  string,  when 
its  ends  are  fastened  at  two  points, 

A and  B,  nearer  together  than  the 
length  of  the  string.  Its  equation  is 


V = \ (e*  + e 


Hence, 


% = I (*’"  “ e_:);  * = l 

If  s be  measured  from  the  lowest  point  V,  to  any  point  P 
(x,  y ),  we  have 

S = \J'J  + e"  •) dx  = I ~ r •)• 


178.  The  Involute  of  a Circle. — (See  Art.  124.)  Let 

0 be  the  centre  of  the  circle, 
whose  radius  is  r ; APR  is  a 
portion  of  the  involute,  T and 
T'  are  two  consecutive  points 
of  the  circle,  P and  Q two 
consecutive  points  of  the  in- 
volute, and  <(>  the  angle  ACT. 

Then  TCT'  = PTQ  = d<p, 
and  PT  - - AT  = rep. 

ds  = PQ  — repdep ; 

s = r J* <pd<p  — lr<t>2  + C. 

If  the  curve  be  estimated  from  A,  (7  = 0,  and  we  have 
s = h~(p2. 

Por  one  circumference,  <p  = 2n  ; s = \r  (27t)?  = 2 rv1. 

For  n circumferences,  <p  — 2nn’  s — %r(2nn)2  = 2rn2n2. 


TEE  CARDIOIDE. 


353 


179.  Rectification  in  Polar  Co-ordinates. — If  the 

curve  be  referred  to  polar  co-ordinates,  we  have  (Art.  102), 

ds2  = r2dd2  + dr2', 


hence  we  get 


or 


180.  The  Spiral  of  Archimedes.- -From  r = ad,  we 

have 

dd  _ 1 
dr  ~ a 


8 — - f (r2  + a2)v  dr 

a J0 

_ r(a*  + »*H  + a 

2 a ' 2 JO& 


4-  V a2  + r 2 


-)• 


(see  Art.  172),  from  which  it  follows  that  the  length  of  any 
arc  of  the  Spiral  of  Archimedes,  measured  from  the  pole,  is 
equal  to  that  of  a parabola  measured  from  its  vertex,  r and 
a having  the  same  numerical  values  as  y andjo. 


181.  The  Cardioide. — The  equation  of  this  curve  is 
r = «(1  + cos  d). 
dr 


Here 


dd 


= — a sin  6, 


and  hence  s = J\*  (i  + cos  d)2  + a 2 sin2  0]*  dd 


— a f (2  + 2 cos  6)^  dd 
= 2a  J cos  ^ dd  = 4a  sin  ^ + C. 


354 


LENGTHS  OF  CURVES  IN  SPACE. 


If  we  estimate  the  arc  s 
from  the  point  A,  for  which 
0 = 0,  we  have 

5 = 0;  .\C=0. 

Making  0 = n for  the 
superior  limit,  we  have 

5 = 4a  sin  - = 4a, 

/V 

which  is  the  length  of  the  arc  ABO ; hence  the  whole 
perimeter  is  8a. 

182.  Lengths  of  Curves  in  Space. — The  length  of 
an  infinitesimal  element  of  a curve  in  space,  whether  plane 
or  of  double  curvature,  from  the  principles  of  Solid  Geom- 
etry (see  Anal.  Geom.,  Art.  169)  is  easily  seen  to  be 

V dx2  + dif  + dz2.* 

Hence,  if  s denote  the  length  of  the  curve,  measured  from 
some  fixed  point  up  to  any  point  P ( x , y,  z ),  we  have 


$ = J v dx2  + dy2  -f-  dz2 


Fig.  47. 


= /[1  + 


! n 

+y 


dx. 


If  the  equations  of  the  curve  are  given  in  the  form 


y - f(x)  and  z = <p  (x), 


we  may  find  the  values  of  ^ and  ^ in  terms  of  x,  and 
J dx  dx 


then  by  integration  s is  known  in  terms  of  x. 


* The  student  who  wants  further  demonstration  of  this,  is  referred  to  Price’s 
Cal.,  Vol.  I,  Art.  341,  and  Vol.  II.  Art.  164;  De  Morgan’s  Dif.  and  Integral  Cal_ 
p.  444 ; and  Homersham  Cox’s  Integral  Cal..  D.  95. 


EXAMPLES. 


355 


183.  The  Intersection  of  Cycloidal  and  Parabolic 
Cylinders. — To  find  the  length  of  the  curve  formed  by  the 
intersection  of  two  right  cylinders,  of  which  one  has  its 
generating  lines  parallel  to  the  axis  of  z and  stands  on  a 
parabola  in  the  plane  of  xy,  and  the  other  has  its  generating 
lines  parallel  to  the  axis  of  y and  stands  on  a cycloid  in  the 
plane  of  xz,  the  equations  of  the  curve  of  intersection  being 

y2  — 4:px,  z = a vers-1  - + V 2 ax  — x2. 


„ dy  p , dz  2a— x 

Here  — \ / - and  \ ; 

dx  V x dx  V x 

- ds  = (:  + | ^ ~ i)2  dx  = (p  + 2 a)i  ~ 

Estimating  the  curve  from  the  origin  to  any  point  P,  we 
have 

s = f (p  -f  2a)i  ^ = 2 (p  + 2 a)^  \/x. 
v 0 x ^ 


EXAMPLES. 

1.  Rectify  the  hypocycloid  whose  equation  is 

1.2  2 
X*  + ys  = 

Ans.  The  whole  length  of  the  curve  is  6a. 

x 

2.  Rectify  the  logarithmic  curve  y = bea. 


Ans.  s = a 


log — 

a + Va2  + y2 


+ v Cl2  + y2  + C. 


3.  Rectify  the  curve 
x — 1 and  x — 2. 


* _ * ±1 
' ~ <r-l 


between  the  limits 


Ans.  s = log  ( e -4-  e-1). 


356 


EXAMPLES. 


4.  Rectify  the  evolute  of  the  ellipse,  its  equation  being 

Put  x = a cos3  9,  y — (3  sin3  9 ; 

then  dx  — —3 a cos 2 9 sin  9 d9, 

dy  — 3/3  sin2  9 cos  9 d9  ; 

s — 3 f («2  cos2  9 + /32  sin2  0)v  sin  9 cos  9 d9 

3 />l*/a?  + 0*  , «2  — /32 

±J0  \ 2 


+ 


cos  29 ) ~ d cos  26 


ft3  — |33 . 

= _ (32  ’ 

KS  — (P 

therefore  the  whole  length  is  4 — -• 

— p i 

If  (3  = a,  this  result  becomes  6 «,  which  agrees  with  that 
given  in  Ex.  1.  (See  Price’s  Calculus,  Vol.  II,  p.  203.) 

5.  Find  the  length  of  the  arc  of  the  parabola  x?  + y*  = a'~ 
between  the  co-ordinate  axes. 


dd 


Put  x = a cos4  9,  y — a sin4  9 ; 

s = 4u  / (cos4  9 + sin4  0)^  sin  9 cos  9 
J o 

= ^ / 1 ( 1 -f  cos2  20)^  d cos  20 

V2J0 

= a + --  log  (V2  + 1). 
v 2 

6.  Find  the  length,  measured  from  the  origin,  of  the 

curve  / y\ 

ic2  = a2(  1 — eaJ. 


A i (O,  -f-  X\ 

Ans.  s = a log  I — x. 

a — xf 


EXAMPLES. 


357 


7.  Rectify  the  logarithmic  spiral  log  r — 9 between  the 

limits  r0  and  r,.  Ans.  s = (1  + (r,  — r0). 

8.  If  100  yards  of  cord  be  wound  in  a single  coil  upon  an 
upright  post  an  inch  in  diameter,  what  time  will  it  take  a 
man  to  unwind  it,  by  holding  one  end  in  his  hand  and 
traveling  around  the  post  so  as  to  keep  the  cord  continually 
tight,  supposing  he  walks  4 miles  per  hour ; and  what  is  the 
length  of  the  path  that  the  man  walks  over? 

Ans.  Time  — 51^  hours;  distance  ==  204T6T  miles. 


9.  Find  the  length  of  the  tractrix  or  equitangential 
curve. 


If  AB  is  a curve  such  that  PT, 
the  length  of  the  intercepted 
tangent  between  the  point  of 
contact  and  the  axis  of  x,  is 
always  equal  to  OA,  then  the 
locus  of  P is  the  equitangential 
curve. 

Let  P and  Q be  two  consecu- 
tive points  on  the  curve ; let 
(x,  y ) be  the  point  P,  and  OA  = 


Y 


PQ  _ a 

PR  ~y’ 

ds  _ a 

dy  ~ ~ y 


(the  minus  sign  being  taken  since  yis  a decreasing  function 
of  s or  x). 


Hence,  s = — a f = 

da  y 


log  y 


— a log 


B58 


EXAMPLES. 


This  example  furnishes  an  instance  of  our  being  able  to 
determine  the  length  of  a curve  from  a geometric  property 
of  the  curve,  without  previously  finding  its  equation. 


The  equation  of  the  tractrix  may  be  found  as  follows  * 


hence 


PR  _ PM 
RQ  ~ MT’ 
dy  _ y 

dx  V«2  - f ’ 


x — 


-j: 


y Va2  — y2 

y 

f‘J  a2dy 

d a a 2 — 


dy 


r 


ydy 

V«2  — y% 


— a log 


a + V a2  — ?/2 


y 


(«2  - fp- 


(See  Ex.  17,  Art.  146.) 


This  curve  is  sometimes  considered  as  generated  by  attaching  one 
end  of  a string  of  constant  length  (=  a ) to  a weight  at  A,  and  by 
moving  the  other  end  of  the  string  along  OX  ; the  weight  is  supposed 
to  trace  out  the  curve,  and  hence  arises  the  name  Tractrix  or  Tractory. 
This  mode  of  generation  is  incorrect,  unless  we  also  suppose  the  fric- 
tion produced  by  traction  to  be  infinitely  great,  so  that  the  weight 
momentum  which  is  caused  by  its  motion  may  be  instantly  destroyed. 
Price’s  Calculus,  Vol.  I,  p.  315. 


10.  A fox  started  from  a certain  point  and  ran  due  east 
300  yards,  when  it  was  overtaken  by  a hound  that  started 
from  a point  100  yards  due  north  of  the  fox’s  starting-point, 
and  ran  directly  towards  the  fox  throughout  the  race.  Find 
the  length  of  the  curve  described  by  the  hound,  both  having 
started  at  the  same  instant,  and  running  with  a uniform 
velocity.  Ans.  354. 13S1  yards. 

This  example,  like  the  preceding,  may  be  solved  without  finding 
the  equation  of  the  curve. 


EXAMPLES. 


359 


11.  Find  the  length  of  the  helix,  estimating  it  from  the 
plane  xy,  its  equations  being 

x = a cos  <p,  y — a sin  <p,  z = cep. 

Ans.  s — ( a 2 -f  c2)~(f). 

12.  Find  the  length,  measured  from  (p  — 0,  of  the  curve 
which  is  represented  by  the  equations 

x — (2a  — 1))  sin  <p  — (a  — b)  sin3  </>, 
y = (2b  — a)  cos  4>  — (b  — a)  cos3  $ . 

Ans.  s = l (a  + b)<p  + | (a  — b)  sin  <p  cos  <f>. 

13.  Find  the  length  of  the  curve  of  intersection  of  the 
elliptic  cylinder  a2y 2 + b2x2  = a2b2,  with  the  sphere 

a2  + y2  + z2  = a2. 

Ans.  2na 


CHAPTER  VII. 

AREAS  OF  PLANE  CURVES. 


184.  Areas  of  Curves. — Let  PM  and  QN  be  two  con- 
Becutive  ordinates  of  the  curve  AB,  and  let  ( x , y)  be  the 
point  P ; let  A denote  the  area  included 
between  the  curve,  the  axis  of  x,  and 
two  ordinates  at  a finite  distance  apart. 

Then  the  area  of  the  trapezoid  MPQN 
is  an  infinitesimal  element  whose  breadth 
is  dx  and  whose  parallel  sides  are  y and 
y + dy ; therefore  we  have 


dA  = 


y + (y  + dy) 
2 


dx  — ydx, 


since  the  last  term,  being  a differential  of  the  second 
order,  must  be  dropped. 


the  integration  being  taken  within  proper  limits.  If,  for 
example,  we  want  the  area  between  the  two  ordinates 
whose  abscissas  are  a and  b,  where  a > £,  we  have 

A = r 'ydx.  (1) 

b 


In  like  manner,  if  the  area  were  included  between  the 
curve,  the  axis  of  y,  and  two  abscissas  at  a finite  distance 
apart,  we  would  have 

A = f xdy, 

’■  d 

where  c and  d are  the  ^/-limits. 


(2) 


QUADRATURE  OF  THE  CIRCLE. 


361 


185.  Area  between  Two  Curves.— If  the  area  were 
included  between  the  two  curves  AB  and  ab,  whose  equa- 
tions are  respectively  y — f(x)  and 
y = (p  ( x ),  and  two  ordinates  CD  and 
EH,  where  OD  = b and  OH  = a, 
we  should  find  by  a similar  course  of 
reasoning, 

A = f L/»  — 0 (s)]  dx. 

The  determination  of  the  area  of  a curve  is  called  its 
Quadrature. 


186.  The  Circle. — The  equation  of  the  circle  referred  to 
its  centre  as  origin,  is  y2  — a2  — x2 ; therefore  the  area  of 
a quadrant  is  represented  by 


A — / (a2  — x2)i' 
0 


dx 


tx  (a2  — x2)?  a 2 . . xia  /0 

— - — - — + — sin-1  - (See  Ex.  4,  Art.  151.) 

/v  & #J() 


__  a^r  * 

~ ~T  ’ 

therefore  the  area  of  the  circle  = ~a2. 


This  result  is  also  evident  from  geometric  considerations 
16 


362 


QUADRATURE  OF  THE  PARABOLA. 


jC  i 

for  the  area  of  the  triangle  OMD  = - (a2  — x2)?,  and  the 

d 2 % 

area  of  the  sector  ODB  = — sin-1  -• 

2 a 


Remark. — The  student  will  perceive  that  in  integrating  between 
the  limits  x = 0 and  x = a,  we  take  in  every  elementary  slice  PQRX 
in  the  quadrant  ADBO ; also  integrating  between  the  limits  x — 0 
and  x — x = OM,  we  take  in  every  elementary  slice  between  OB 
and  MD.* 


187.  The  Parabola. — From  y 2 — 2 px, 
we  have 

y = V 2px. 

Hence,  for  the  area  of  the  part  OPM, 
we  have 

A = V%P  i x?dx  — -f  V 2p  x? ; i.  e.,  \xy. 
J o 


Fig.  52. 


Therefore  the  area  of  the  segment  POP', 

cut  off  by  a chord  perpendicular  to  the  axis,  is  f of  the 

rectangle  PHH'P'. 


188.  The  Cycloid. — From  the  equation 


= r vers-1  - — V 2 ry  — y2, 


we  have 


dx  = 


ydy 


V 2 ry  — y2 

a = rr  y2dy— 

' o v 2 ry  — y2 


— frrr2. 


(See  Ex.  6,  Art.  151)  = 4 the  area  of  the  cycloid.  Since 
integrating  between  the  limits  includes  half  the  area  of 
the  figure. 


* The  student  should  nay  close  attention  in  every  case  to  the  limits  of  the 
integration. 


AREA  BETWEEN  PARABOLA  AND  CIRCLE.  363 


Therefore  the  whole  area  = 3~r2,  or  three  times  the  area 
vf  the  generating  circle .* 

189.  The  Ellipse.— The  equation  of  the  ellipse  referred 
to  its  centre  as  origin,  is 

a2y1  _|_  l)2x2  — a2gi  . 

therefore  the  area  of  a quadrant  is  represented  by 

A = - f\a 2 - .t2)’  dx 
aJ  o 

= ^ (See  Art.  186)  r=  ^abn. 

Therefore  the  area  of  the  entire  ellipse  is  ~ab. 


2 — 


190.  The  Area  between  the  Parabola  y 
and  the  Circle  y2  — 2*ax  — oc2.  — These  curves 
through  the  origin,  and  also  intersect  at 
the  points  A and  B,  whose  common  abscissa 
is  a.  Hence,  to  find  the  area  included 
oetween  the  two  curves  on  the  positive  side 
of  the  axis  of  x,  we  must  integrate  between 
the  limits  x — 0 and  x — a.  Therefore, 
by  Art.  185,  we  have 


ax 

pass 


A = f \(2ax  — x 2)^  — (a:r)5]  dx 


o 

TT  cd 

T 


(See  Ex.  6,  Art.  151.) 
which  is  the  area  of  OPAP'. 


Fig.  53. 


* This  quadrature  was  first  discovered  by  Eoberval,  one  of  the  most  distin- 
guished geometers  of  his  day.  Galileo,  having  failed  in  obtaining  the  quadrature 
by  geometric  methods,  attempted  to  solve  the  problem  by  weighing  the  area  of  the 
curve  against  that  of  the  generating  circle,  and  arrived  at  the  conclusion  that  the 
former  area  was  nearly,  but  not  exactly,  three  times  the  latter.  About  1628, 
Roberval  attacked  it,  but  failed  to  solve  it.  After  studying  the  ancient  Geometry 
for  six  years,  he  renewed  the  attack  and  effected  a solution  in  1634  (See  Salmon’! 
Higher  Plane  Curves,  p.  266.) 


364 


THE  SPIRAL  OF  ARCHIMEDES. 


191.  Area  in  Polar  Co-ordinates. — Let  the  curve  be 
referred  to  polar  co-ordinates,  0 being  the  pole,  and  let 
OP  and  OQ  be  consecutive 
radii-vectores,  and  PR  an 
arc  of  a circle  described  with 
0 as  centre ; let  ( r , 6)  be  the 
point  P.  Then  the  area  of  the 
infinitesimal  element  OPQ 
= OPR  -f  PRQ ; but  PRQ  is 

an  infinitesimal  of  the  second  order  in  comparison  with 
OPR,  when  P and  Q are  infinitely  near  points ; conse- 


quently the  elementary  area  OPQ  = area  OPR  = 


r2dd 


Hence  if  A represents  the  area  included  between  the  curve, 
the  radius-vector  OP,  and  the  radius-vector  OB  drawn  to 
some  fixed  point  B,  we  have 


= */ 


r2d0. 


If  (3  and  « are  the  values  of  6 corresponding  to  the  points 
B and  C respectively,  we  have 

A = l far2dd. 

Q 

192.  The  Spiral  of  Archimedes.  — Let  r = ^ be 

its  equation.  Then 

A = tt  f A dr  = ^-r3  -f-  C. 

If  we  estimate  the  area  from  the  pole,  we  have  A = C 
when  r = 0,  and  (7=0;  hence, 


A = inr*. 


which  is  the  value  of  the  area  passed  over  by  the  radius- 
vector  in  its  revolution  from  its  starting  at  c to  any  value. 


as  r. 


EXAMPLES. 


365 


If  we  made  6 = 2n,  we  have  r = 1 ; therefore 
A = 

which  is  the  area  described  by  one  revolution  of  the  radius- 
vector.  Hence  the  area  of  the  first  spire  is  equal  to  one- 
third  the  area  of  the  measuring*  circle. 

If  we  make  9 = 2 (2tt),  r = 2 ; therefore 

A = frr, 

which  is  the  whole  area  described  by  the  radius-vector 
during  two  revolutions,  and  evidently  includes  twice  the 
first  spire  + the  second.  Hence  the  area  of  the  first  two 
spires  = |tt  — — ftr,  and  so  on. 

EXAMPLES. 

1.  Find  the  area  of  y = x — .r3  between  the  curve  and 

the  axis  of  x.  Ans. 

The  limits  will  be  found  to  be  x — 0,  x = + 1 ; also  x = 0, 
x = — l.f 

2.  Find  the  area  of  y = x3  — b2x  between  the  curve 

and  the  axis  of  x.  Ans.  f /A 

3.  Find  the  area  of  y = x3  — ax2  between  the  curve  and 

the  axis  of  x.  Ans.  ^a4. 

4.  Find  the  whole  area  of  the  two  loops  of  a2y2 

= x 2 ( a 2 — x 2).  Ans.  -fa2. 

5.  Find  the  area  of  xy2  = a3  between  the  limits  y = h 

and  y = c.  . 0 „ 5 — c 

J Ans.  2 a3 — y — 

be 

6.  Find  the  whole  area  of  the  two  loops  of  Ay2 

= a2b2x2  — b2xA.  Ans.  $ab. 


* See  Anal.  Geom.,  Art.  15S. 

t The  student  should  draw  the  figure  in  every  case,  and  determine  the  limits  of 
the  integrations. 


366 


EXAMPLES. 


7.  Find  the  whole  area  of  a2y2  = x3  (2 a — x).  (See 
Arts.  150  and  188.)  Ans.  va2. 


8.  Find  the  whole  area  between  the  Cissoid  y 3 — 
and  its  asymptote.  (See  Art.  103.) 


cc3 


2a  — x 
Ans.  ?>-dl. 


9.  The  equation  of  the  hyperbola  is  a2y2  — lAx2  = — aW ; 

find  the  area  included  between  the  curve,  the  axis  of  x,  and 
an  ordinate.  Xy  j /x  _|_  y' xi  _ a2. 

Ans.  -f-j  log  ( s )• 

10.  The  equation  of  the  Witch  of  Agnesi  is 

= 4«2  (2«  — y) ; 

find  the  area  included  between  the  curve  and  its  asymptote. 

Ans.  4a2~. 

11.  Find  the  area  of  the  catenary  VPMO,  Fig.  45. 

Ans.  (e*  — e a^j  = a (y2  — a2)K 

12.  Find  the  area  of  the  oval  of  the  parabola  of  the  third 
degree  whose  equation  is  cy2  = (x  — a)  (x  — b)2.  (See 

Art.  142.)  . 8 ,,  b 

Ans.  — — — \b  — ap. 
l°yc 

13.  Find  the  area  of  one  loop  of  the  curve 

ay2  = x2  (a2  — x2)?. 

14.  Find  the  whole  area  between  the  curve 

x2y2  + aW  = a2y 2 

and  its  asymptotes.  Ans.  2tt ab. 

15.  Find  the  whole  area  of  the  curve 


Ans.  fa2. 


(f)?  = 


Ans.  ab. 


EXAMPLES. 


367 


16.  Find  the  area  included  between  the  parabola  = 2px 
and  the  right  line  y = ax. 

These  two  loci  intersect  at  the  origin  and  at  the  point  whose  ab 


17.  Find  ( 1 ) the  area  included  between  the  parabola 


the  right  line  passing  through  the  focus  and  inclined  at  45c 
to  the  axis  of  x,  and  the  left-hand  double  ordinate  of  inter- 
section. (See  Art.  185.) 

Also  find  (2)  the  whole  area  between  the  line  and  parabola. 

(1.)  Here  the  r-limits  are  found  tobe^  (v2  + l)  an<l  ^ (\/2— 1 Y 
hence  we  have 


= -Lip2  — 3y dip2  + \/2 p2  = p2  — 2\/2),  Ans. 

(2.)  Ans.  |pV2. 

18.  Find  the  whole  area  included  between  the  four  infi- 


ll). Find  the  area  of  the  Naperian  logarithmic  spiral. 

Ans.  \r2. 

20.  Find  the  whole  area  of  the  Lemniscate  r2  = a2  cos  29. 

Ans.  a2. 


scissa  is 


— : hence  the  z-limits  are  0 and  ™ ; therefore,  Art  185, 


y2  — 2px, 


nite  branches  of  the  tractrix. 


Ans.  Tra2. 


368 


EXAMPLES. 


21.  Find  the  whole  area  of  the  curve 

r — a (cos  26  + sin  26). 

Ans.  -na2. 

22.  Find  the  area  of  the  Cardioide.  (See  Art.  181.) 

A ns.  | ~a2. 

23.  Find  the  area  of  a loop  of  the  curve  r = a cos  n6. 

- a 2 
4n 


An?. 


24.  Find  the  area  of  a loop  of  the  curve 
r — a cos  nd  -f  l sin  nd. 


Ans. 


a2  + V1  rt 


25.  Find  the  area  of  the  thrrfc  loops  of  the  curve 
r — a sin  36.  (See  Fig.  33.) 

Ans. 


TTQr 


26.  Find  the  area  included  between  the  involute  and  tht 
evolute  in  Fig.  46,  when  the  raring  has  made  one  revolution. 

Ans.  irW. 


CHAPTER  VIII 


AKEAS  OF  CURVED  SURFACES. 

193.  Surfaces  of  Revolution. — If  any  plane  be  sup- 
posed to  revolve  around  a fixed  line  in  it,  eveiy  point  in  the 
plane  will  describe  a circle,  and  any  curve  lying  in  the  plane 
will  generate  a surface.  Such  a surface  is  called  a surface 
of  revolution ; and  the  fixed  line,  round  which  the  revolu- 
tion takes  place,  is  called  the  axis  of  revolution. 

Let  P and  Q be  two  consecutive 
points  on  the  curve  AB ; let  ( x , y ) be 
the  point  P,  and  s the  length  of  the 
curve  AP  measured  from  a fixed  point 
A to  any  point  P.  Then  MP  = y, 

1STQ  = y + dy,  and  PQ  ==  ds. 

Denote  by  S the  area  of  the  surface 
generated  by  the  revolution  of  AP 
around  the  axis  OX;  then  the  surface  generated  by  the 
revolution  of  PQ  around  the  axis  of  x is  an  infinitesimal 
element  of  the  whole  surface,  and  is  the  convex  surface  of 
the  frustum  of  a cone,  the  circumferences  of  whose  bases 
are  2 Try  and  2n  ( y -f  dy),  and  whose  slant  height  is  PQ  = ds  ; 
therefore  we  have 

ds  = ^+.2;At#)  pQ  = 

since  the  last  term,  being  an  infinitesimal  of  the  second 
order,  must  be  dropped.  Therefore,  for  the  whole  surface, 
we  have 


S = 2n  J* yds  — 2n  J* y\/ dx2  -f-  dy2, 


370 


QUADRATURE  OF  THE  SPHERE. 


the  integral  being  taken  between  proper  limits.  If  for 
example,  we  want  the  surface  generated  by  the  curve  be- 
tween the  two  ordinates  whose  abscissas  are  a and  b,  where 
a > b,  we  have 


In  like  manner  it  may  be  shown  that  to  find  the  surface 
generated  by  revolving  the  curve  round  the  axis  of  y , we 
have 


194.  The  Sphere. — From  the  equation  of  the  gener- 
ating curve,  x2  + y2  = r2,  we  have 


Hence,  the  surface  of  the  zone  included  between  two 
planes  corresponding  to  the  abscissas  a and  b is 


that  is,  the  area  of  the  zone  is  the  product  of  the  circum- 
ference of  a great  circle  by  the  height  of  the  zone. 

To  find  the  surface  of  the  whole  sphere,  we  integrate 
between  + r and  — r for  the  ^-limits ; hence  we  have 


that  is,  the  whole  surface  of  the  sphere  is  four  times  the 
area  of  a great  circle. 

Remark.— If  a cylinder  be  circumscribed  about  a sphere,  its  convex 
surface  is  equal  to  2~r  x 2 r — 4 -r4,  which  is  the  same  as  the  surface 
of  the  sphere.  If  we  add  2-;-2  to  this,  which  is  the  sum  of  the  areas 
of  the  two  bases,  we  shall  have  for  the  ivhole  surface  cf  ‘he  cylinder 


y — ( r 2 — x 2)v  and 

a v ' dx  y* 

S — 2n  j* y(l  + dx  = 2n ■ J*\ rdx  — 2nrx  + C. 


b 


— r 


QUADRATURE  OF  PARABOLOID  OF  REVOLUTION.  371 


6 Hence  the  whole  surface  of  the  cylinder  is  to  the  surface  of  the 
sphere  as  3 is  to  2.  This  relation  between  the  surfaces  of  these  two 
bodies,  and  also  the  same  relation  between  the  volumes,  was  discovered 
by  Archimedes,  who  thought  so  much  of  the  discovery  that  he  ex- 
pressed  a wish  to  have  for  the  device  on  his  tombstone,  a sphere 
inscribed  in  a cylinder.  Archimedes  was  killed  by  the  soldiers  of 
Marcellus,  B.  C.  212,  though  contrary  to  the  orders  of  that  general. 
The  great  geometer  was  buried  with  honors  by  Marcellus,  and  the 
device  of  the  sphere  and  cylinder  was  executed  upon  the  tomb.  140 
years  afterward,  when  Cicero  was  questor  in  Sicily,  he  found  the 
monument  of  Archimedes,  in  the  shape  of  a small  pillar,  and  showed 
it  to  the  Syracusans,  who  did  not  know  it  was  in  being , he  says  it  was 
marked  with  the  figure  of  a sphere  inscribed  in  a cylinder.  The 
sepulchre  was  almost  overrun  with  thorns  and  briars.  See  article 
“ Marcellus,”  in  Plutarch’s  Lives,  Vol  III,  p.  120. 


195.  The  Paraboloid  of  Revolution.  — From  the 
equation  of  the  generating  curve  y 2 = 2 px,  we  have 


which  is  the  surface  generated  by  the  revolution  of  the 
part  of  the  parabola  between  its  vertex  and  the  point 

(*»  y)- 

We  might  have  found  the  surface  in  terms  of  y instead 
of  x,  as  follows  : 


dx  _y^ 
dy  p 


373 


THE  PROLATE  SPHEROID. 


= :ip  i(f  + - pzl 


which  result  agrees  with  (1),  as  the  student  can  easilj 
verify. 

196.  The  Prolate  Spheroid  (See  Anal.  Geom.,  Art, 
191). — From  the  equation  of  the  generating  curve 

f — (1  — e2)  ( a 2 — x2), 

we  have 


therefore  for  half  the  surface  of  the  ellipsoid,  since  the 
^-limits  are  a and  0,  we  have 


2n  yds  = 2n  Vl  — e2  V a2  — x2  ds 


= 2n  Vl  — e2  V a2  — e2x?  dx  (Art.  175.) 


2n  be 
a 


(See  Ex.  4,  Art.  151.) 


= tos[(i  - o*  + 


= nb2  -1 sin-1  e. 


e 


197.  The  Catenary. — From  the  equation  of  the  gen- 
erating curve, 


SURFACE  GENERATED  BY  THE  CYCLOID. 


373 


we  have  for  the  surface  of  revolution  around  the  axis  of  x 
between  the  limits  x and  0, 


rtx  nx  / x _f\ 

S — 2n  J yds  — -na  J + e “j 


ds 


x\2 


= inaf%  + e dx  (by  Art.  177) 

= ?[!(—)+**] 

= 77  [f  (e~a  ~ e~)  + ax\ 

= n (ys  + ax),  (where  s = VP,  Fig.  45.) 

198.  The  Surface  generated  by  the  Cycloid  when 
it  revolves  around  its  axis. — From  its  equation 


y — r vers-1  - + V 2rx  — x2, 
J r 


we  have 


dy 

dx 


2r  — x 


x 


* = (*  + M dx 


^ dx . 

x 


df\ h 
dx V 

S = 2n  J yds  = 2n  f y dx. 


(1) 

(2) 

(3) 

(4) 


Put 


u = y,  dv  = \/  ~dx; 


r.  du  — dy,  and  v — 2 V 2rx ; 
therefore  (by  Art.  147)  we  have 


374 


POLAR  CO-ORDINATES. 


J‘ y \J~  dx  — 2y  V 2rx  — 2 V2r  J Vx  dy 

= 2y  V2rx  — 2 V2r  J a/2 r — xdx  [by  (2)] 

= 2 V2rx  (r  vers-1  J-  + *>/2rx  — x^j  + f V2r  (2 r — x)i 

[by  (1)  and  integrating.] 

f(y  V ^ 4Tr/'2  “ 

which  in  (4)  gives 

S = 8 n2r2  — 2£-7rr2  = 8rrr2  (7r  — J). 

199.  Surfaces  of  Revolution  in  Polar  Co-ordi- 
nates.— If  the  surface  is  generated  by  a curve  referred  to 
polar  co-ordinates,  its  area  may  be  determined  as  follows  : 
Let  the  axis  of  revolution  be  the  initial  line  OX,  see 
Fig.  54,  and  from  P (r,  0 ) draw  PM  perpendicular  to  OX. 
Then  PM  = r sin  6,  and  the  infinitesimal  element  PQ 
= ds  will,  in  its  revolution  round  OX,  generate  an  infini- 
tesimal element  of  the  whole  surface,  whose  breadth  = ds 
and  whose  circumference  = 2nr  sin  6.  Hence, 

S = / 2~r  sin  6 ds * — 2n  J*  r sin  6 ^r2  + 

(Art.  179) 

the  integral  being  taken  between  proper  limits. 

200.  The  Cardioide. — From  Art.  181,  we  have 

Q 

ds  = a ( 2 + 2 cos  6 )?  dd  — 2 a cos  - dd. 


* This  expression  might  have  been  obtained  at  once  by  substituting  in  Art.  193, 
for  y,  its  value  r sin  6. 


D 0 UBLE  INTEGRA  TION. 


375 


For  the  surface  of  revolution  of  the  whole  curve  about 
the  initial  line,  we  have  n and  0 for  the  limits  of  6,  there- 
fore we  have 

S — j 2 nr  sin  6 ds 
do 

p*  6 

= 4 ~a2  / (1  + cos  6)  cos  - sin  6 dd 
<7  o 2 

Pn  6.6 

= 16tj a2  / cos4  - sm  - d6 
«/  0 ” * 

— zj-Tra2  cos5  f j = 2£na2. 

5 2J0  5 

201.  Any  Curved  Surfaces. — Double  Integration. — 

Let  (x,  y,  z)  and  ( x + dx,  y -\-  dy,  z -f  dz)  be  two  consecu- 
tive points  p and  q on  the  sur- 
face. Through  p let  planes  be 
drawn  parallel  to  the  two  planes 
xz  and  yz  ; also  through  q let 
two  other  planes  be  drawn  par- 
allel respectively  to  the  first. 

These  planes  will  intercept  an 
infinitesimal  element  pq  of  the 
curved  surface,  and  the  projec- 
tion of  this  element  on  the 
plane  of  xy  will  be  the  infini- 
tesimal rectangle  PQ,  which  = dx  dy. 

Let  S represent  the  required  area  of  the  whole  surface, 
and  dS  the  area  of  the  infinitesimal  element  pq,  and 
denote  by  a,  (3,  y,  the  direction  angles*  of  the  normal  at 
p (x,  y,  z).  Then,  since  the  projection  of  dS  on  the 
plane  of  xy  is  the  rectangle  PQ  = dx  dy,  we  have  by  Anal. 
Georn.,  Art.  168, 

dx  dy  = dS  cos  y.  (1) 


z 


* See  Anal.  Geom.,  Ait.  170. 


376 


SURFACE  OF  A SPHERE. 


Similarly,  if  dS  is  projected  on  the  planes  yz  and.  zx, 
we  have 

dy  dz  = dS  cos  a ; (2) 

dz  dx  = (IS  cos  (3.  (3) 


Squaring  (1),  (2)  and  (3),  and  adding,  and  extracting 
the  square  root,  we  have 

dS  = {dxldy2  -f-  dy^dz2  + dz2dx1)^ 

(since  cos2  « + cos2  (3  + cos2  y = 1, 

Anal.  Geom.,  Art.  170). 

S — J*  J*  ( dx2dy 2 -f-  difdz2  + dz2dx2)? 


the  limits  of  the  integration  depending  upon  the  portion 
of  the  surface  considered. 


202.  The  Surface  of  the  Eighth  Part  of  a Sphere.— 

Let  the  surface  represented  in  Fig.  56  be  that  of  the 
octant  of  a sphere  ; then  0 being  its  centre,  its  equation  is 


Hence, 


x2  + y2  + = «2. 


dz  _ x dz  y 

dx  2 ’ dy  z 


Now  since  pq  is  the  element  of  the  surface,  the  effect 
of  a ^/-integration,  x being  constant,  will  be  to  sum  up 
all  the  elements  similar  to  pq  from  H to  1;  that  is, 
from  y — 0 to  y — 'Ll  = ?/,  = v/a2  — a? : and  the  aggre- 


EXAMPLES. 


377 


gate  of  these  elements  is  the  strip  ~E.pl.  The  effect  of  a 
subsequent  ^-integration  will  be  to  sum  all  these  elemental 
strips  that  are  comprised  in  the  surface  of  which  OAB 
is  the  projection,  and  the  limits  of  this  latter  integra- 
tion must  be  x = 0 and  x — OA  = a.  Therefore, 
we  have 


8 = 


a dx  dy 


o V a2 


r?  — 


a dx  dy 
o d0  V.yi'  — y2 


= fa\adx  sin-i  y-T. 
do  •-  ydo 

7 TO2 

= yo  i”adx  = T 


EXAMPLES. 

1.  Find  the  convex  surface  of  a right  circular  cone, 
whose  generating  line  is  ay  — bx  — 0. 

A ns.  nb  V a2  + t/2. 

Remark. — It  is  evident  that  the  projection  of  the  convex  sur- 
face of  a right  circular  cone  on  the  plane  of  its  base,  is  equal 
to  the  base;  hence  it  follows  (Anal.  Geom.,  Art.  168)  that  the 
convex  surface  of  a right  circular  cone  is  equal  to  the  area  of  its 
base  multiplied  by  the  secant  of  the  angle  between  the  slant 
height  and  the  base.  Thus,  calling  this  angle  a,  we  have  in 
the  above  example, 

S = ttS2  sec  « = 7ri5  = nb  + 6!, 

which  agrees  with  the  answer. 

2.  Find  the  area  of  the  surface  generated  by  the  revolu- 
tion of  a logarithmic  curve,  y = ex,  about  the  axis  of  x , 
between  the  -^-limits  0 and  y. 

Ans.  n [y  (1  + y2)i  + log  [?/  + (!  + y2)*]  \. 


378 


EXAMPLES. 


3.  Find  the  area  of  the  surface  generated  by  the  revolu- 
tion of  the  cycloid  (i)  about  its  base,  and  (2)  about  the 
tangent  at  the  highest  point. 

Ans.  (1)  hf-TTo2 ; (2)  ^na2. 


4.  Find  the  area  of  the  surface  generated  by  the  revolu- 
tion of  the  catenary  about  the  axis  of  y,  between  the 
2-limits  0 and  x.  Ans.  2 n \xs  — a{y  — a)]. 


By  Art.  177, 


S = 2-n-  xds  = 2n  | xs  — f scf:rj  , 


from  which  we  soon  obtain  the  answer. 


5.  Find  the  area  of  the  surface  of  a spherical  sector,  the 
vertical  angle  being  2 a and  the  radius  of  the  sphere  = r. 

Ans.  4 ~r2  ^sin  ” j • 

6.  Find  the  area  of  the  surface  generated  by  the  revolu- 

tion of  a loop  of  the  lemniscate  about  its  axis,  the  equation 
being  r2  = a 2 cos  26.  Ans%  na2  (2  _ 2i). 

Here  find  rds  = a^dd  ; etc. 

7.  Find  the  area  of  the  surface  generated  by  the  revolu- 

tion of  a loop  of  the  lemniscate  about  its  axis,  the  equation 
being  r 2 = a2  sin  26.  Ans.  2~a2. 

8.  A sphere  is  cut  by  a right  circular  cylinder,  the  radius 
of  whose  base  is  half  that  of  the  sphere,  and  one  of  whose 
edges  passes  through  the  centre  of  the  sphere.  Find  the 
area  of  the  surface  of  the  sphere  intercepted  by  the  cylinder. 

Let  the  cylinder  be  perpendicular  to  the  plane  of  x y ; 
then  the  equations  of  the  cylinder  and  the  sphere  are 
respectively  y2  = ax  — x2  and  x2  + y2  + z2  = a2.  It  is 
easily  seen  that  the  ^/-limits  are  0 and  V ax  — x2  = and 
the  2-limits  are  0 and  a.  Therefore,  Art.  201,  we  have 


EXAMPLES. 


37 


S 


r>a  nyx 

J a <J  n 


a dx  dy 


o u o V a2  — x 2 — y 2 
j (ax  — x2)i 
(a2  - x 2)* ' 


= a C sin  1 — dx 

do 


= af,°sin~'  + x) 


= «[(«  + *)  sin- 

= ^(i-4 


(Art.  147) 


Therefore,  the  whole  surface  = 2a2  (rr  — 2).  (In  Price’s 
Calculus,  Vol.  II,  p.  326,  the  answer  to  this  example  is 
d 3 (tt  — 2),  which  is  evidently  only  half  of  what  it  should 
be.) 


9.  In  the  last  example,  find  the  area  of  the  surface  of  the 
cylinder  intercepted  by  the  sphere. 

Eliminating  y,  we  have  2 = V a2  — ax  for  the  equation 
of  the  projection  on  the  plane  xz  of  the  intersection  of  the 
sphere  and  the  cylinder.  Therefore  the  2-limits  are  0 and 
2,  = V a2  — ax,  and  the  z-limits  are  0 and  a;  hence,  Art. 
201,  we  have 


(dx2  dy 2 + dy2  dz 2 + dz 2 dx2~fi  = 


1 + 


dxdz 


a dx  dz 
2\/  ax  — x2 


for  an  element  of  the  surface  of  the  cylinder. 


dxdz 
\/ ax  — x2 


therefore  the  whole  area  of  the  intercepted  surface  of  the 
cylinder  is  4a2.  (See  Gregory’s  Examples,  p.  436.) 


10.  The  axes  of  two  equal  right  circular  cylinders  inter- 
sect at  right  angles  ; find  the  area  of  the  one  which  is  inter- 
cepted by  the  other.  Ans.  8 a2. 


380 


EXAMPLES. 


Let  the  axes  of  the  two  cylinders  be  taken  as  the  axes  of  y and  z 
and  let  a — the  radius  of  each  cylinder.  Then  the  equations  are 

a2  + s2  = a2,  x-  + y*  — a?. 

11.  A sphere  is  pierced  perpendicularly  to  the  plane  of 
one  of  its  great  circles  by  two  right  cylinders,  of  which  the 
diameters  are  equal  to  the  radius  of  the  sphere  and  the  axes 
pass  through  the  middle  points  of  two  radii  that  compose  a 
diameter  of  this  great  circle.  Find  the  surface  of  that  por- 
tion of  the  sphere  not  included  within  the  cylinders. 

Ans.  Twice  the  square  of  the  diameter  of  the  sphere. 

12.  Find  the  area  of  the  surface  generated  by  the  revolu- 
tion of  the  tractrix  round  the  axis  of  x.  Ans. 

13.  If  a right  circular  cone  stand  on  an  ellipse,  show  that 
the  convex  surface  of  the  cone  is 

5 (OA  + OA')  (OA-OA')*  sin  «, 

where  0 is  the  vertex  of  the  cone,  A and  A'  the  extremities 
of  the  major  axis  of  the  ellipse,  and  a is  the  semi-angle  of 
the  cone  at  the  vertex.  (See  Remark  to  Ex.  1.) 


CHAPTER  IX. 


VOLUMES  OF  SOLIDS. 


203.  Solids  of  Revolution. — Let  the  curve  AB,  Fig.  55, 
revolve  round  the  axis  of  x,  and  let  V denote  the  volume 
of  the  solid  bounded  by  the  surface  generated  by  the  curve 
and  by  two  planes  perpendicular  to  the  axis  of  x,  one 
through  A and  the  other  through  P ; then  as  MP  and  NQ 
are  consecutive  ordinates,  the  volume  generated  by  the  revo- 
lution of  MPQH  round  the  axis  of  x is  an  infinitesimal 
element  of  the  whole  volume,  and  is  the  frustum  of  a cone, 
the  circumferences  of  whose  bases  are  2iry  and  2n  (y  + dy), 
and  whose  altitude  is  MX  = dx ; therefore  we  have 

d V = dx  = nfdx, 

o 


by  omitting  infinitesimals  of  the  second  order.  Hence,  for 
the  whole  volume  generated  by  the  area  between  the  two 
ordinates  whose  abscissas  are  a and  b,  where  a > b,  we 
have 

pa 

V — / ny2dx. 
d'6 

In  like  manner,  it  may  be  shown  that  to  find  the  volume 
generated  by  revolving  the  arc  round  the  axis  of  y,  we  have 


204.  The  Sphere. — Taking  the  origin  at  the  centre  of 
the  sphere,  we  have  y2  = a2  — x2 ; therefore  we  have 


V = n 


— x2)  dx 


tt  ( a2x  — %x?) 


a 

—a 


= i^3, 


for  the  whole  volume  of  the  sphere. 


382 


VOLUME  GENERATED  BY  CYCLOID. 


Cor.  1. — To  find  the  volume  of  a spherical  segment  be- 
tween two  parallel  planes,  let  b and  c represent  the  distances 
of  these  planes  from  the  centre ; then  we  have 


V — tt  ( a 2 — x 2)  dx  = n \a2  ( b — c)  — \ ( b 3 — e3)]. 


Cor.  2. — To  find  the  volume  of  a spherical  segment  with 
one  base,  let  li  be  the  altitude  of  the  segment ; then  b = a 
and  c — a — h,  and  we  have 


Cor.  3.  %na3  — § of  nd 2 x 2a  — f of  the  circumscribed 
cylinder.  (See  Art.  194,  Remark.) 

205.  The  Volume  generated  by  the  Revolution  of 
the  Cycloid  about  its  Base. 


and  integrating  between  the  limits  y — 0 and  y — 2r,  we 
find  for  the  whole  volume 

r=2n  r & 

L'o  \/2ry  — y2 

= | r / — -U!jL=  (by  Ex.  6,  Art.  151) 

Jo  y2ry  — y2 

= ig-rn  (|/  27t)  (by  Ex.  6,  Art.  151) 

_ 5~2/.3. 

We  have  5TT2r3  = -fv  (2r)2  x 2n r. 

Hence,  the  volume  generated  by  the  revolution  of 
the  cycloid  about  its  base  is  equal  to  five-eighths  the 
circumscribing  cylinder. 


V — — j‘  (a2  — a:2!  dx  — —Id  ( a — ■ 


Here 


dx  = ■ — — ^ (Art.  176) ; 


V 2 ry  — y2 


SOLIDS  BOUNDED  BY  ANY  CURVED  SURFACE.  383 


206.  The  Cissoid  when  it  revolves  round  its 
Asymptote. — Here  OM  = x,  MP  = y, 

OA  = 2a,  MA  — 2a  — x,  HD  = dy  ; 
hence  an  infinitesimal  element  of  the 
whole  volume  is  generated  by  the  revo- 
lution of  PQDH  about  AT,  and  is 
represented  by  n (2a  — x)~  dy. 

The  equation  of  the  Cissoid  is 

✓y»3 


f - 


dy  = 


2a  — x 

(3 a — x)  (2 ax  — x 2)J 


dx ; 


(2a  — x)2 

hence,  between  the  limits  x — 0 and  x — 2a,  we  have 

/>  2a  p 2a 

V—2tc  (2  a — xf  dy  — 2tt  (3a—x)(2ax — x2)* 
do  do 

i2a  6a2x  — 5 ax2  + a;3 


dx 


— 2n  / 
do 


V 2 ax  — x2 


■ dx  — 2n 2 a3 


(by  Ex.  6,  Art.  151). 


207.  Volume  of  Solids  bounded  by  any  Curved 
Surface. — Let  (x,  y,  z)  and 
(x  + dx,  y + dy,  z-\-dz)  be  two 
consecutive  points  E and  F 
within  the  space  whose  volume 
ig  to  be  found.  Through  E 
pass  three  planes  parallel  to 
the  co-ordinate  planes  xy,  yz, 
and  zx;  also  through  F pass 
three  planes  parallel  to  the 
first.  The  solid  included  by 
these  six  planes  is  an  infinitesi- 
mal rectangular  parallelopipe- 
don,  of  which  E and  F are  two  opposite  angles,  and  the 
volume  is  dxdy  dz  ; the  aggregate  of  all  these  solids  between 


384 


TRIPLE  INTEGRATION. 


the  limits  assigned  by  the  problem  is  the  required  volume. 
Hence,  if  V denote  the  required  volume,  we  have 


the  integral  being  taken  between  proper  limits. 

In  considering  the  effects  of  these  successive  integrations, 
let  us  suppose  that  we  want  the  volume  in  Fig.  58  contained 
within  the  three  co-ordinate  planes. 

The  effect  of  the  ^-integration,  x and  y remaining  con- 
stant, is  the  determination  of  the  volume  of  an  infinitesimal 
prismatic  column,  whose  base  is  dxdy,  and  whose  altitude 
is  given  by  the  equations  of  the  bounding  surfaces  ; thus,  in 
Fig.  58,  if  the  equation  of  the  surface  is  z =f(x,y),  the 
limits  of  the  z-integration  ar ef(x,  y ) and  0,  and  the  volume 
of  the  prismatic  column  whose  height  is  P p is  / (x,y)  dxdy, 
hence  the  integral  expressing  the  volume  is  now  a double 
integral  and  of  the  form 


If  we  now  integrate  with  respect  to  y,  x remaining  con- 
stant, we  sum  up  the  prismatic  columns  which  form  the 
elemental  slice  H plmq,  contained  between  two  planes  per- 
pendicular to  the  axis  of  x,  and  at  an  infinitesimal  distance 
(dx)  apart.  The  limits  of  y are  JJ  and  0,  L l being  the  y to 
the  trace  of  the  surface  on  the  plane  of  xy,  and  which  may 
therefore  be  found  in  terms  of  x by  putting  z = 0 in  the 
equation  of  the  surface  ; or,  if  the  volume  is  included  be- 
tween two  planes  parallel  to  that  of  xz,  and  at  distances  ya 
and  yx  from  it,  y0  and  y,  being  constants,  they  are  in  that 
case  the  limits  of  y;  in  the  same  way  we  find  the  limits  if 
the  bounding  surface  is  a cylinder  whose  generating  lines 
are  parallel  to  the  axis  of  z.  In  each  of  these  cases  the 
result  of  the  ^-integration  is  the  volume  of  a slice  included 
between  two  planes  at  an  infinitesimal  distance  apart,  the 
length  of  which,  measured  parallel  to  the  axis  of  y,  is  a 


EXAMPLES. 


385 


function  of  its  distance  from  the  plane  of  yz\  thus  the  limits 
of  the  ^-integration  may  be  functions  of  x,  and  we  shall 
have 

V — J ' j f(x,  y)  clx  dy  — J F(x)  dx, 

where  F(x ) dx  is  the  infinitesimal  slice  perpendicular  to 
the  axis  of  a;  at  a distance  x from  the  origin,  and  the  sum 
of  all  such  infinitesimal  slices  taken  between  the  assigned 
limits  is  the  volume.  Thus,  if  the  volume  in  Fig.  58  be- 
tween the  three  co-ordinate  planes  is  required,  and  OA  = a, 
then  the  .r-limits  are  a and  0.  If  the  volume  contained 
between  two  planes  at  distances  x0  and  xl  from  the  plane  of 
yz  is  required,  then  the  a>limits  are  x0  and  xx. 


EXAMPLES. 


1.  The  ellipsoid  whose  equation  is 


a?  y 2 z 2 

a2  + P + ? ~ lm 


l — — 2 — | -j 

and  0,  which  call  zx  and  0 ; the  limits  of  y are  J2  = 

/ 2/2\  -g- 

1)\1  — —2)  and  0,  which  call  yx  and  0 ; the  a:-limits  are  a 
and  0. 


First  integrate  with  respect  to  z,  and  we  obtain  the  infini- 
tesimal prismatic  column  whose  base  is  PQ,  Fig.  58,  and 
whose  height  is  P p.  Then  we  integrate  with  respect  to  y, 
and  obtain  the  sum  of  all  the  columns  which  form  the 
elemental  slice  H plmq.  Then  integrating  wuth  respect  to  x, 
we  obtain  the  sum  of  all  the  slices  included  in  the  solid 
OABC. 

pa  pyx  pzx 

...  V = 8 j J J dx  dy  dz 

17 


386 


EXAMPLES. 


f*  a r*  y \ / /) /2\  i 

= 8 c / (1  — - — f-)  dx  dy 

J0  \ «2  b2/  3 

= jj  yo  yo  («/i2  - «/2)i  % 

-UT!W“^+ 

_ 8c  f‘ayli 

- b Jo  2 2 

= J*  (a2  — x 2)  dx  = §nabc, 


y?  . ,y 
sin-1  - 
2 v 


dx 


2.  The  volume  of  a right  elliptic  cylinder  whose  axis 
coincides  with  the  axis  of  x and  whose  altitude  = 2a,  the 
equation  of  the  base  being 

c2y2  + b2z2  — Vi?. 

Here  the  2-limits  are  | ( l 2 — y2)^  and  0,  which  call  zx  and 
0 ; the  ^-limits  are  b and  0 ; the  .r-limits  are  a and  0. 


F=8 


dx  dy  dz 


= f\v-f$dxdy 


rhrr  / 

= 8 — — / dx  — 2 abcn. 

(See  Price’s  Calculus,  Yol.  II,  p.  356.) 


S.  The  volume  of  the  solid  cut  from  the  cylinder  x2  yz 
= a 2 by  the  planes  z — 0 and  z — x tan  «. 

Here  the  2-limits  are  x tan  a and  0,  or  2,  and  0;  the 
^-limits  are  (a2  — x2)?  and  — (a2  — x2) v,  or  yx  and  — yx ; 
the  x-limits  are  a and  0. 


EXAMPLES. 


387 


pa  py1  pzx 

V — / / dx  dy  dz 

Jo  J-y^Jo 


-ta  pyx 


( x tan  «)  dx  dy 


= 2 tan  a f x (a3  — x 2)^  dx  = § a3 


tan 


The  volume  of  the  solid  common  to  the  ellipsoid 
p + p + p = 1 and  the  cylinder  a;2  + i/2  = £2. 

yy2\  ^ 

l — — — j 

and  0,  or  z,  and  0 ; the  limits  of  the  ^'-integration  are 
(62  — y2)\  and  0,  or  x1  and  0;  the  ^/-limits  are  0 and  b.* 

pb  pXi  />z, 

V = 8 / / / dy  dx  dz 

Jo  J o J o 


= */y 

0 ^ Q 


6 p ~~ 


x 2 y2 


dy  dx 


MH 


i a2~TalJ2 


dy 

o 


__  /** 

- V,/0  |8V-  P /'  3 ~ apt)1 

= (J<  _ ~ - (**  ~ 2ft]‘ 

= + (i  - 1!)  i]  * 
= ^ + si,1"‘  ll A1  - »)** 


* In  this  example,  this  order  of  integration  is  simpler  than  it  would  he  to  take 

'♦  with  respect  to  y and  then  x. 


388 


EXAMPLES. 


= iHC[^  («2  ~ ^ + Sin_1 

(See  Mathematical  Visitor,  1878,  p.  26.) 

208.  Mixed  System  of  Co-ordinates. — Instead  of 
dividing  a solid  into  columns  standing  on  rectangular 
bases,  so  that  z dx  dy  is  the 
volume  of  the  infinitesimal 
column,  it  is  sometimes  more 
convenient  to  divide  it  into 
infinitesimal  columns  standing 
on  the  polar  element  of  area 
abed  — r dr  dO,  in  which  case 
the  corresponding  parallelopipedon  is  represented  by 
zr  dr  dO,  and  the  expression  for  V becomes 

y=J'f  zr  dr  dO, 

taken  between  proper  limits.  From  the  equation  of  the 
surface,  z must  be  expressed  as  a function  of  r and  0. 


c 


EXAMPLES. 


1.  Find  the  volume  included  between  the  plane  z — 0, 
and  the  surfaces  x2  + y2  — 4 az  and  y2  — 2 cx  — x2. 

ofi  -I—  V2  T2  T2 

Here  z = — y— = — : hence  the  z-limits  are  — and  0. 
4 a 4 a 4a 

The  equation  of  the  circle  y2  = 2 cx  — x2,  in  polar  co-or- 
dinates, is  r = 2c  cos  0 ; hence  the  r-limits  are  0 and 

2c  cos  0,  or  0 and  r j ; and  the  0-limits  are  0 and  ^ • 


.-.  V = 


2 


dO  dr 


9 /4  r*  In  9 (A 

— — / cos4  Odd  — — jSgW.  (Ex.  4,  Art.  157.) 
a Po  a 

_ 3 7 TC4 

8 a 


EXAMPLES. 


389 


2.  The  axis  of  a right  circular  cylinder  of  radius  b, 
passes  through  the  centre  of  a sphere  of  radius  a,  when 
a > b ; find  the  volume  of  the  solid  common  to  both 
surfaces.* 

Take  the  centre  of  the  sphere  as  origin,  and  the  axis  of 
the  cylinder  as  the  axis  of  z;  then  the  equations  of  the 
surfaces  are  x2  + y2  + z2  = a2  and  x2  + y2  = b2\  or,  in 
terms  of  polar  co-ordinates,  the  equation  of  the  cylinder 
is  r — b. 

Hence  for  the  volume  in  the  first  octant,  the  2-limits  are 
sj a* — x2  — y2  or  V a2  — r2  and  0;  the  r-limits  are  b 

. 7T 

and  0 ; the  0-linnts  are  - and  0. 


= y [a3  - ( a 2 - a*)t]. 

(See  Gregory’s  Examples,  p.  428.) 

209.  The  polar  element  of  plane  area  is  r dr  d0  (Art. 
208).  Let  this  element  revolve  round  the  initial  line 
through  the  angle  2n,  it  will  generate  a solid  ring  whose 
volume  is  2 nr  sin  Or  dr  dd,  since  2 nr  sin  8 is  the  circumfer- 
ence of  the  circle  described  by  the  point  (r,  6).  Let  <p 
denote  the  angle  which  the  plane  of  the  element  in  any 
position  makes  with  the  initial  position  of  the  plane  ; 
then  d(p  is  the  angle  which  the  plane  in  any  position  makes 


* This  example,  as  well  as  the  preceding  one,  might  he  integrated  directly  in 
terms  of  x and  y by  the  method  of  Art.  207,  but  the  operation  would  be  more  com- 
plex than  the  one  adopted. 


390 


EXAMPLES. 


with  its  consecutive  position.  The  part  of  the  solid  ring 
which  is  intercepted  between  the  revolving  plane  in  these 
two  consecutive  positions,  is  to  the  whole  ring  in  the  same 
proportion  as  d(f>  is  to  2 n.  Hence  the  volume  of  this 
intercepted  part  is 


which  is  therefore  an  expression  in  polar  co-ordinates  for 
an  infinitesimal  element  of  any  solid.  Hence,  for  the 
volume  of  the  whole  solid  we  have 


in  which  the  limits  of  the  integration  must  be  so  taken 
as  to  include  all  the  elements  of  the  proposed  solid.  In 
this  formula  r denotes  the  distance  of  any  point  from  the 
origin,  6 denotes  the  angle  which  this  distance  makes  with 
some  fixed  right  line  through  the  origin  (the  initial  line), 
and  (p  denotes  the  angle  which  the  plane  passing  through 
this  distance  and  the  initial  line  makes  with  some  fixed 
plane  passing  through  the  initial  line.  (See  Lacroix  Cal- 
cul  Integral,  Vol.  II,  p.  209.) 

The  order  in  which  the  integrations  are  to  be  effected  is 
theoretically  arbitrary,  but  in  most  cases  the  form  of  the 
equations  of  surfaces  makes  it  most  convenient  to  integrate 
first  with  respect  to  r ; but  the  order  in  which  the  6-  and 
^-integrations  are  effected  is  arbitrary. 


1.  The  volume  of  the  octant  of  a sphere.  Let  a = the 
radius  of  the  sphere  ; then  the  limits  of  r are  0 and  a ; 
hence. 


In  thus  integrating  with  respect  to  r,  we  collect  all  the 
elements  like  r2  sin  d d(p  dO  dr  which  compose  a pyramidal 


r2  sin  6 dcp  dO  dr, 


EXAMPLES. 


EXAMPLES. 


32: 


solid,  having  its  vertex  at  the  centre  of  the  sphere,  and  foi 
its  base  the  curvilinear  element  of  spherical  surface  which 
is  denoted  by  a 2 sin  6 d<f>  dO. 

Integrating  next  with  respect  to  6 between  the  limits  0 


In  thus  integrating  with  respect  to  6,  we  collect  all  the 


shaped  slice  of  the  solid  contained  between  two  consecutive 
planes  through  the  initial  line. 


In  this  example  the  order  of  the  integrations  is  imma- 
terial. 

2.  The  volume  of  the  solid  common  to  a sphere  of 
radius  a,  and  the  right  circular  cone  whose  vertical  angle 
is  2 a and  whose  vertex  is  at  the  centre  of  the  sphere. 

Here  the  -/'-limits  are  0 and  a,  the  0-limits  are  0 and  <&. 
the  0-limits  are  0 and  2m 


and  x , we  have 
z 


v = ft  [<- cos  *>]>  = /r 


71 

Lastly,  integrating  with  respect  to  0 from  0 to  ^ 
we  have 


V =—■  (See  Todhunter’s  Int.  Cal.,  p.  183.) 


o 


g-  (1  — cos  a)  dtp 


— | no?  (1  — cos  d). 


392 


EXAMPLES, 


EXAMPLES. 


1.  Find  the  volume  of  a paraboloid  of  revolution  whose 
altitude  = a and  the  radius  of  whose  base  = b. 

Ans.  - alp' 

2 

2.  Find  the  volume  of  the  prolate  spheroid.  Also  of 
the  oblate  spheroid.  Ans.  The  prolate  spheroid  = ab2. 

The  oblate  spheroid  = %TUilb. 

3.  Find  the  volume  of  the  solid  generated  by  the  revolu- 
tion of  y — a*  about  the  axis  of  x,  between  the  limits  x 
and  — co , where  a > 1. 


Ans.  - a 


,2x 


(log  a)- 


4.  Find  the  volume  of  the  solid  generated  by  the  revolu- 

tion of  y = a log  x about  the  axis  of  x,  between  the 
limits  x and  0.  Ans.  -ncPx  (log2  x — 2 log  x + 2). 

5.  Find  the  volume  of  the  solid  generated  by  the 
revolution  of  the  tractrix  round  the  axis  of  x.  Ans.  f '«3. 

6.  Find  the  volume  of  the  solid  generated  by  the 
revolution  of  the  catenary  round  the  axis  of  x. 

Tt 

Ans.  jr  a (ys  -j-  ax).  (Compare  with  Art.  197.) 

Ai 


7.  Find  the  volume  generated  by  the  revolution  of  a 

parabola  about  its  base  2b,  the  height  being  h*  (See 
Art.  206.)  \§^bh\ 

8.  The  equation  of  the  Witch  of  Agnesi  being 


x 2 = 4a2 


2a  — y 
V ’ 


find  the  volume  of  the  solid  generated  by  its  revolution 
round  the  asymptote.  Ans.  4 *2as. 


* This  solid  is  called  a parabolic  spindle. 


tsX AMPLE S. 


393 


8.  Find  the  volume  of  a rectangular  parallelopipedonj 
three  of  whose  edges  meeting  at  a point  are  a,  b,  c.  (See 
Art.  207.)  Ans.  abc. 


10.  Find  the  volume  contained  within  the  surface  of  an 
elliptic  paraboloid*  whose  equation  is 

f 

a 


+ — 2a;, 


and  a plane  parallel  to  that  of  yz,  and  at  a distance  c from  it. 

Ans.  -nc2(ab)^. 

11.  The  axes  of  two  equal  right  circular  cylinders  inter- 
sect at  right  angles,  their  equations  being  x2  + z2  = a2  and 
x2  -f  y2  — a2)  find  the  volume  of  the  solid  common  to  both. 

Ans.  ^6-a3. 

12.  A paraboloid  of  revolution  is  pierced  by  a right  cir- 

cular cylinder,  the  axis  of  which  passes  through  the  focus 
and  cuts  the  axis  of  the  paraboloid  at  right  angles,  the 
radius  of  the  cylinder  being  one-fourth  the  latus-rectum  of 
the  generating  parabola;  find  the  volume  of  the  solid  com- 
mon to  the  two  surfaces.  . 3 /2  tt\ 

Ans- P \ 3+^/- 

Here  tlie  equations  of  the  surfaces  are 

yi  + z-=2px  and  x^  + y-  = px. 

13.  Find  the  volume  of  the  solid  cut  from  the  cylinder 
x3  y2  = 2 ax  by  the  planes  z — x tan  « and  z~x  tan  /3. 

Txci 3 

Ans.  2 (tan  j3  — tan  «)  • 

A 

14.  Find  the  volume  of  the  solid  common  to  both  sur- 
faces in  Ex.  8 of  Art.  202.  (See  Art.  208.) 

Ans.  § (3 n — 4)  a3. 

15.  Find  the  volume  of  the  part  of  the  hemisphere  in  the 
last  example,  which  is  not  comprised  in  the  cylinder. 

Ans.  fa3. 


* Called  elliptic  ’paraboloid  because  the  sections  made  by  planes  parallel  to  the 
planes  of  xy  and  xz  are  parabolas,  while  those  parallel  to  the  plane  of  yz  are 
ellipses.  (Salmon’s  Anal.  Geom.  of  Three  Dimensions,  p.  58.) 


394 


EXAMPLES. 


16.  Find  the  volume  of  the  solid  intercepted  between  the 

concave  surface  of  the  sphere  and  the  convex  surface  of  the 
cylinder  in  Art.  208,  Ex.  2.  Ans.  fr.  (fla  _ l2)l. 

17.  Find  the  volume  of  the  solid  comprised  between  the 

22-fva 

surface  z = ae  ca  and  the  plane  of  xy.  Ans.  ~a&. 
Here  the  r-limits  are  0 and  oo  ; and  the  ^-limits  are  0 and  2~. 


18.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  the  cardioide  r — a (1  + cos  9)  about  the  initial 
line. 


n n /?  r » 

Here  V=  / 

t/o  ''0  ^0 

(See  Art.  209.) 


a (1+cosfl) 

r%  sin  9 dd  d(p  dr  — etc. 


Ans. 


8vfl3 


19.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  the  Spiral  of  Archimedes,  r = ad,  about  the  initial 
line  between  the  limits  9 = n and  9 = 0. 

Ans.  1 7T2a3  (~2  — G). 

20.  A right  circitiar  cone  whose  vertical  angle  = 2 «,  has 
its  vertex  on  the  surface  of  a sphere  of  radius  a,  and  its 
axis  coincident  with  the  diameter  of  the  sphere  ; find  the 
volume  common  to  the  cone  and  the  sphere. 

47ra3 


Ans. 


(1  — cos4  «). 


21.  Find  the  volume  of  a chip  cut  at  an  angle  of  45°  to 

the  centre  of  a round  log  with  radius  r.  (Mathematical 
Visitor,  1880,  p.  100.)  Ans.  p-3. 

22.  Find  the  volume  bounded  by  the  surface 


©*+  (!)*+  ©'  = 


and  the  positive  sides  of  rhe  three  co-ordinate  planes. 


Ans. 


abc 

90" 


EXAMPLES. 


395 


23.  Find  the  volume  of  the  solid  bounded  by  the  tnree 
surfaces  x2  + y2  = cz,  x2  + if  = ax,  and  z = 0. 


24,  A paraboloid  of  revolution  and  a right  circular  cone 
nave  the  same  base,  axis,  and  vertex,  and  a sphere  is 
described  upon  this  axis  as  diameter.  Show  that  the  volume 
intercepted  between  the  paraboloid  and  cone  bears  the  same 
ratio  to  the  volume  of  the  sphere  that  the  latus-rectum  of 
the  parabola  bears  to  the  diameter  of  the  sphere. 

25.  Find  the  volume  included  between  a right  circular 
cone  whose  vertical  angle  is  60°  and  a sphere  of  radius  r 
touching  it  along  a circle,  by  the  formula 


20.  In  the  right  circular  cone  given  in  Ex.  13  of  Art 
802,  prove  that  its  volume  is  represented  by 


dx  dy  dz. 


nr° 


\ (OA«OA')*  sin3  a cos^- 
6 


MISCELLANEOUS  EXAMPLES. 


Nearly  all  of  the  following  examples  have  been  collected 
by  me  and  given  to  my  students  during  the  last  twenty 
years  of  my  teaching.  They  are  here  arranged  so  as  to 
illustrate  and  enforce  every  part  of  the  subject.  The 
examples  in  each  chapter  come  under  the  corresponding 
chapter  in  the  book.  No  principle  is  well  learned  by  a 
pupil  until  he  can  use  it;  and  for  this  purpose  he  must 
have  much  practice  in  solving  examples.  The  numerous 
examples  here  given  are  for  the  convenience  of  the  instruc- 
tor, that  he  may  have,  year  by  year,  a sufficient  variety 
from  which  to  select. 


CHAPTER  II. 


Find  ^ 
ax 


in  each  of  the  following  examples : 


Vl  — x2 

(1  + x2)i ' 


Ans. 


2.  y = (x2  + 1)  Vx3  — x. 


3. 

4. 

5. 


y = 
y = 


y = 


2 Vx 

3 + x2 

x 


Va2  - x2 

1 + 4x2  j-z q 2 

Uwr-Vl-2x- 


3x3  + 2 
x (x3  + l)f  ’ 


2x  (2  — .x2) 

V(1  — x2)  (1  + x2)5 
7x4  - 2x2  - 1 
2 Vx 3 — x 
3 (1  - x2) 

(3  + x2)“Vx 

a2 

(a2  — x2)^ 

1 

x 4 Vl  — 2x2 
— 2 

x2  (x3  + 1)3 


396 


MISCELLANEOUS  EXAMPLES. 


397 


7.  y = log 


8.  y = log 


Vl  + ar  -f-  Vl  — a;2 


>•  y = log  y - 


X V2  + Vl  + X 
Vl  — a;2 

Vl  + ar*  + a; 


x Vl  — a:4 
V2 

(1  - a;2)  Vl  + X2 

1 


V 1 + x2  — a;  Vl  + a;2 

10.  y = -^(log2x  + 41ogx  + 8).  — - ■ 

Vx  a:? 


Vl  + x . * /V 1 + X — 1 

— +IogVvrrITT 


Ans.  — 


c2Vl  + 


12.  y - 


-(£ 


6x6  24  x4 


— o i + vi  — ar 
■x-+r6l°g ; 


Ans.  — 


C7  ViW 


13  y = x — + 24  log  x 

* V 2x2  4 (2x  - l)2  + S 

45  8x6  — 1 

-Ti»g(a.-i). 

14.  y = (e*  — e~xf  (e2x  + 2e4x  + 3e6x). 

Aws.  24e6x  (e2*  — 1). 

2\lT  x2  - 1 

3T2  — 4 

— 3 tan2  x sin3  x. 

1 

sin4  x cos2  x 


i''-' IV 


15. 

16.  y = sec  x(sin4x  + 4 sin2x  — 8). 
sec  x 4 cos  x 


17.  y 


sm3x 


— | cot  X. 

3 sm3  x 3 


18.  y = 2 tani  x (1  + - tan2  x). 

5 

19.  y = tan4  x — 2 tan2  x + log  see4  x. 


1 

sini x cosl x 
4 tan5  x. 


398 


MISCELLANEOUS  EXAMPLES. 


20.  y 

21.  y 

22.  y 

23.  y 

24.  y 

25.  i/  = 


sec  x + log  tan  - 

jJ 


tan3  x 


+ tan  x. 


vers  ^ + xj  vers  ^ — x 
tan  £ + § tan3  x + | tan5  x. 
log  tan  x + tan2  x + \ tan4  x. 


cos  x/1  3 \ 3 . ^ x 

^~\sln4_x  + 2 sin2xj~  8l0g  tan2' 

Ans.  — 


sm  x cos2  x 
sec4  x. 

— sin  2x. 

sec6  x. 

1 

sin  x cos5  x 


X 

26.  y = J sec3  x + sec  x + log  tan  ^ • 

27.  y = log  — X ^ --  + V3  tan-1 


sm°  x 
1 


sm  x cos4  x 


28.  y 

29.  y 


Vx2  + x + 1 

- 1\*  V2 


-1os(^tt) 


Ans. 

x 


+ tan-1  - 
3 V2 


2x  + 1 

V3 

3x 

x3  — 1 

x2 


e0*  sm  x 


x4  + x2 
26“* 


30.  y = cos 

31.  y 


(a  sin  x — 2 cos  x)  + ; ■ 

2 4a  + a3 

Ans.  e“*  sin2  x. 

4 


4 + a2 

3 + 5 cos  x 


5 + 3 cos  x 


. 1 + x V2  + x2 

lOg*  — — 

1 — x V2  + x2 


2 tan- 


5+3  cos  x 
x V2 


Ans. 


1 -x2 
4 V2 


1 + x4 


32.  y = log  — + lo 

1 — X 


’g\/ 


'1  + X + X2 


1 — X + x: 


/oi  x V3 
— + Vo  tan  ^ — — -§• 
:2  1 — x2 


6 


Ans. 


1 — x‘ 


MISCELLAXEO  US  EXAMPLES. 


399 


33.  y 

34.  y 

35.  y 

36.  y 

37.  y 

38.  y 

39.  y 

40.  y 

41.  y 

42.  y 

43.  y 

44.  y 


= 4=  tan-1  X 

V2  Vl  - x2 


x3  — 3x 
2 Vl  — x 


/ 1 


= — = tan-1  (— — tan 


V3 
= log 


W3 
(x  - l)i  1 


(i  + x2)  vr 

x4 

(1  - x2)f 

1 

2 + cos  x 


tan 


(x2+2)*  3V2  V2 


Ans. 


= sin-1 


— tan-1 


x tan  a 


,Va2  — x 

2x 


1 + 3x2 


+ cot-1 


x3  — x2  + 2x  — 2 
a2  tan  a 

(a2  — x2)  Va2  — x2  sec2  a 
+ tan-1 2x. 


Ans. 


1 + 2x2 
3 

1 + 9x2 


log  \ \ 


_ 9 


^X 


sin' 


tan 


sec- 


2x2  + 2x  + 1 
Ans. 

6 + a cos  x 
a + 6 cos  x 
L Va2  — 62  sin  x 
fe  + a cos  x 

1 


+ tan-1 


2x 


1 - 2x2 
8x2 


4x4  + 1 

Va2  - 52 
a + b cos  x 
Va2  - b2 
a + b cos  x 
2 


tan 

1 


2x2  - 1 

_x  Vl  + x2  — 1 . 


Vl-X2 


2 (1  + x2) 


lo£ 


■ V2  + 1 


4V2  x2  + x V2  + 1 2V2 


+ tan  1 (x  V2  — 1)]. 


Ans. 


[tan_1(x  V2  + 1) 
x2 


1 + x4 


400 


MISCELLANEOUS  EXAMPLES. 


Vx4  — 


45.  y = - log 

6 x2  + 1 


2 V3 

— tan-1  (2x  + V3)].  A ns. 


[tan-1  (2x  — V3) 
x3 


1 + x6 


46.  v = 


sin5  x 


8 


COS3  X + 7T  COS  X — 


cos  x 


3 . 


+ 


3x 

128 


^ sin3  x + 2 sm  x 
Ans.  sin4  x cos4  x. 


CHAPTER  IV. 


Prove  the  following  : 

1.  y = (x2  — 6x  + 12)  e*. 

2.  = 

3.  y = -g-  cos  x cos3  x« 

4.  y = (x2  — 3x  + 3) 


*)■ 


Ans.  f'"(x)  = xV. 

/"  (x)  = x log  x. 

f"'(x)  = sin3  x. 
/'"(x)  = 8x2e2x. 


5.  y = tan2  x + 8 log  cos  x + 3x2.  /"  (x)  = 6 tan4  x. 

6.  y = x3(3  log2  x — 11  log  x + V')-  f”'(x)  *=  18  log2  x. 

Ans.  /"  (x)  = ±- 


7.  n2=  :• 

2a  -x 

a:  x 

8-  y = \ («“  + e a)  . 

9.  x = a vers-1  - — ^J2  ay  — 


3 a2 


10.  y = y/ — — — . 

T x — a 

11.  2/  = e2VA 

12.  y = x*. 


Vx  (2a  — x)5 
3a2 

4 Vx  (x  — a)5 
(2  Vx  — 1) 

2x  Vx 

/"(x)  = x1  (1  + logx)2  + x*-:t. 


Ans.  f"(x)  = 


r\x)  = 


MISCELLANEOUS  EXAMPLES. 


401 


13.  y = T^— . 
y 1 - x 

14.  y = x*  log  (x*). 


/v(z)  = 
/v(*)  = 


120 

24 


, Vl  + x2  — 1 _ 2x 

15.  v = tan-1 h tan  1 t -■ 

a x 1 — x2 


^Ins.  f"{x)  = 


5x 


16.  y = log  \/l±. 


V2  + 


V2  + 


+ tan- 


Ans.  f"{x)  — — 


(1  + x2)2 

■ i s . 

1 - x2 

8 V2x3 


(1  + x4)2 


CHAPTER  V. 

Develop  each  of  the  following  expressions : 

. , x2  4x4  31x6 

6 6 1 [2  ^ [4  [6 


2.  e1  sec  x = 1 + x + x2  + g x3  + y + ^ x5 

3 e*sin*  = i + X2  + + . . . . 


4.  etan  * = 1 + x + 


x3  7x4 
(T  “ 24 


5.  log  (1  + +)  = log  2 + * + £ - * 

x2  2 13 

6.  e*seca:  = 1 +x  + y + gX3  + -^X4  •• 


7.  e 


rV=T. 


12 

x4  / — =-x5 

+g+ 


£ 


402 


MISCELLANEO  US  EXAMPLES. 


rp 2 /y»4  /y.6  *1  7 /y»8 

8.  log  sec  x = y + + 45  + 2520 


9.  e8 


A,  , X‘  x 

-x  = e(l  + ^ + 


2 + 3 


. , , lx2  , 1 • 3x5  , 1 • 3 • 5x7 

10.  Sin-1  x = X + o—o  + O .4  r + o 


2-3  ' 2-4-5  1 2-4- 6-7 

t i . j . x2  2X3  9x3 

11.  e log  (1  + x)  = x + j^-  + -jg-  + • • • • 


12.  cos2 


l/2  , 2^ 

2 \ [2  + [4 


2V 

[6 


CHAPTER  VI. 

Evaluate  tlie  following  functions : 

wlien  x = 0. 


i ex  - e~x 
1-  — ; » 


sm  x 


o loS  (2*2  - 1) 
— : — — > 


3. 


tan  (x  — 1) 

2 tan  x — sin  2x 
sin3  x 


. sec2  x — 2 tan  x 

4-  q — : q > 

1 + cos  4x 

(e*  — e2)3 
(x  - 4)  e1  + e2  x : 


5. 


x = 1. 
x = 0. 

7 r 

X = 4 
x = 2. 


x 


„ sin  x — sm-1  x 

O.  : — q } 

sm3  x 

es*  _ ioe2-+3  + 15^+4  _ 6e5 

• e4*  _ 6e2x+2  + 8e*+3  _ 3g4  ’ x - • 

sin  2x  + 2 sin2  x — 2 sin  x 
cos  x — cos2  x 


8. 


= 0. 


0. 


Ans.  2. 

4. 

2. 

1 

2 

6e4. 


5e 

~2 


MISCELLANEOUS  EXAMPLES. 


'og  (x  — 


tan  x 


10 


. ( ax  — l)x, 


11.  sin  x log  cot  x, 

12.  cosec2  x — -r  , 

xL 

13.  (1  — tan  x)  sec  2x, 


14.  (a2  — x2)  tan 
v ' 2a 

2 1 


15. 


sin2  x 1 — cos  x ' 
16.  2x  tan  x — it  sec  x, 
tan 


17. 

18. 
19. 


1&  + 1) 


tan 


ttX 


(x  — 2)  e*  + x + 2 
(e*  - l)3 

x V3x  — 2x4  — xa/x 


1-x* 


20.  ^a2  — x2  cot 


TTX 


21.  (1  — x)  tan 


e*  + log 


22. 


tan  x — x 


3.  (c 


23.  ^cos  2x)x2 , 


x~2' 

x = co  . 
x = 0. 

x,=  0. 


X 4 


x = a. 


= 0. 


* = 2' 


x = 1. 

x = 0. 

x = 1. 

x = a. 

x = 1. 

x = 0. 


403 


0. 


log  a 

0. 

1 

3 * 

1. 

4a2 

7 r 

1 

2* 

- 2. 

2. 

1 

6 ‘ 

81 
20  ‘ 

4 a 

7T 

2 

IT 


1 


O' 


1 


x = 0. 


404 


MISCELLANEOUS  EXAMPLES. 


tan 


, TTX 

. tan  — 
7 TX\  2 


(e*  + l)*  > 


24.  (cot  x)8m  % 

25. 

26. 

27.  (x  — l)log  sin  nx  } 

28. 

29. 

30. 


TTX\ 


tan  nx 


31. 

32. 

33. 


tan-j 

1 

(cot  x)log*> 

1 

[log  (e  + x)]  *, 
(ex  + x)x, 


^.log  sin  x, 

tan  X 

tan  5x 
tan  x 


tan  3x  ’ 

sec  x 
sec  3x 


x sin  x — „ sec  x, 


34 

35. 

36. 

37. 

38. 

n 

39.  (cos  mx)x\ 


when  x = 0.  Ans. 
x = 1. 

X = oo. 


X = 1. 

X = 1. 

x = 0. 

x = 0. 

x = 0. 
x = 0. 


7T 

* = 2' 


x = 


x = 


IT 

x = r 


2x2  2x  tan  ttx 

(ex  —e~x)2—2x2  (e*  + e"1) 


x = 0. 
> x = 0. 
x=  0. 


1 

e 

e. 

ea. 


1 

e 

i 

e2. 

e°. 

5. 


- f- 

nm2 


MISCELLANEOUS  EXAMPLES. 


405 


40. 


41. 

42. 


tan  x\4 


x 

tan  x\— , 


x 


x = co  1. 

x = 0.  e%. 

2 = 0.  oo. 


CHAPTER  VII. 


In  each  of  the  following  seven  examples  find  the  total 
derivative : 

1 . y , . - . / du\  — 1 

1 m — ran  1 — . nr2  - 


= a2. 

Ans. 

/du\ 

\dx) 

a sin  x, 

2 = 

cos  2. 

= (a2  + 1)  e°* 

sin  2. 

3.  u = sin  - 5 z = ex,  y = x2. 


Am ■ (s)=(i-2)ScosS- 

= sin  y. 

= 2 a Q/  cos  y + sin  ?/)  + cot  y. 


4 . u — 2axy  + log  x,  x — sin  y. 

. (du 
Ans 


dy 

5.  u = y2  + z4  + zy,  y = sin  x,  z = cos  2. 

Ans.  = cos  22  (1  — sin  2x ). 

6.  it  = tan-1  — , 2 = eh  y = e_h 

2 + y y 


Ans.  f^)  = 


2e2* 


dz  ) eiz  + 1 


406 


MISCELLANEOUS  EXAMPLES. 


7.  u = x4y2  — X-^-  + x4,  y = log  x. 
du 


Arts,  (-fo.)  = x3  (4  log2  x + 3£). 


8.  If  u = (x2  + y show  that 


, d?u 
dx 2 


+ 2xj/ 


d2u 


cPu 


dx  dy  + r dy2  °‘ 


9.  If  u = (x3  + y3)K  show  that 


d?ii  9 

zi&  + 2x« 


d?u 


+ y 2 


cPu. 


= -rU. 


10.  If«  = 


(x2  + y2  + z2)  % 


dx  dy  y dy2  4 
, show  that 


cPtt  d2w  _ „ 
dx2  dy2  dz2 

11.  If  w = log  (x3  + y3  + z3  — 3xyz),  show  that 

3 


du  ^ du  ^ du 
dx  dy  dz 


cPu 

dtf 

d2u 

dx2 


d2u 

dy2 

d2u 

W2 


x + y + z 

<Pu  3 

dz2 

ddu 
dz 2 


(x  + y + z)2 


14.  Also 


2 d2u  2 d2u  2d?u 

4 1 1 

dx  dy  dy  dz  dz  dx 

d?u 


(x  + y + z)2 


2 du  du  du  _ 
dx  dy  dz  dx  dy  dz 


6 


x3  + y3  + z3-  3 xyz 
Change  the  independent  variable  from  x to  y in  the  two 
following  equations  : 


dPx  cPx_0 

A dy3  + df~ 


(cPyV 

__  dy 

d3y 

d2y . 

(dy\ 

[dx2 

dx 

dx3 

dxil 

[dx 

MISCELLANEOUS  EXAMPLES. 


407 


16.  (3af  + 2)(ptf=(ad/+l)‘§L  • % 

V dx  ) \dx  7 \ ax  /ax  ax 3 


Ans. 


d2x Y_  + d3x 


dy  Jdy 3 

Change  the  independent  variable  from  x to  z in  the  five 
following  equations  : 

17.  xi  + a2y  = 0,  where  x = - • 

ax 1 z 

dry  2 dy-  , A 

^iS-  M+-zdz  + ay  = °- 

18.  (1  — x2)  ™ — x ~ + a2y  = 0,  where  x = sin  z. 

Ans.  + a%y  = °' 


cPy 


dy 


+ 


4 n2y 


dx2  A ^ e2x  + e 21  dx  1 (e2x  + e 2X)2 


= 0, 


where  x = log  Vtan  z. 


Ans • + n2y  = 0. 


20.  (2x  - l)3  ^ + (2x  - 1)  ^ = 2y,  where  2x  = 1 + e*. 
ax3  dx 

Ans.  4^1-12^1+9^  = !,. 
dz3  dz2  dz 


21.  x4$(  + 6x3  + 9x2^  + 3x^  + y = log  x, 

v3  ax3  ax 


dx4  ' dx3 
and  x = 6s. 


A™.  ^ + 2^  + „=2. 


dz 4 


dz2 


22.  Change  the  independent  variable  from  y to  x in 
^ — 4 tan  y ~ + 2 tan2  y ^ = 0.  and  tan  y = x. 

Ans.  (l+x=)^  + 2,(l+^,^  + 2^  = 0 


\ 


408 


MISCELLANEOUS  EXAMPLES. 


23.  Change  the  independent  variable  from  6 to  x in 


Find  the  values  of  x which  give  maximum  and  mini- 
mum values  of  the  following  functions  : 


Ans.  x = — 1,  a max. ; x = 3,  a min. 


4.  y — 4x3  — 15x2  + 12x  — 1. 

Ans.  x = a max. ; x = 2,  a min. 

5.  y = x3  — 3x2  4-  6x. 


Ans.  There  is  neither  a max.  nor  a min. 


6.  y = 3x5  — 125x3  + 2160x. 

Ans.  x = — 4 and  3,  max. ; 
x = — 3 and  4,  min. 

_ x2  — 7x  + 6 

7-  y=  x — 10  ' 

Ans.  x = 4,  a max. ; x = 16,  a min. 


CHAPTER  VIII. 


1.  y = J x3  — 2x2  + 3x  + 1. 

Ans.  x = 1,  a max. ; x = 3,  a min. 

2.  y = 2x3  — 9x2  + 12x  — 3. 

Ans.  x = 1,  a max. ; x = 2,  a min. 

3.  y = x3  — 3x2  — 9x  + 5. 


8.  y = xx . 

9.  y = 2 tan  x — tan2  x 


Ams.  x = e,  a max. 


x = T j a max. 
4 


7T 


10.  y = sin  x cos  (x  — a). 


a 7r 

x = ^ + -j  > a max 


MISCELLANEOUS  EXAMPLES. 


409 


11.  y = (x-  2)5  (2x  + l)4. 

12.  y 


X ==  S'  J 

x = a min-  5 
x = 2,  neither. 


(x  + If  (x  - 5)2. 

Ans.  x = 5 and  — 1,  min. ; x = a max. 


13.  y = 

14.  y -- 

15.  y 

16.  y 


xi  — x 


+ 1 


Ans. 


x = 0,  a max. ; 
x = 2,  a min. 


x2  + x — 1 

x3  - 9x2  + 24x  + 16. 

Ans.  x = 2,  a max. ; x = 4,  a min. 

: x3  — 2x2  — 4x  + 1. 

Ans.  x = — f,  a max. ; x = 2,  a min 

: 2x3  - 21x2  + 36x  - 20. 

Ans.  x = 1,  a max. ; x = 6,  a min. 

7 r 


17.  y = tan  x + 3 cot  x. 

18.  y = sin  x (1  + cos  x). 

19.  y = x (a  + x)2  (a  — x)3. 

Ans.  x = ^ and  ■ 


20.  y = — + 


b2 


21.  J/  = 


22.  y = 


x a — x 


3x2  - a2 


Ans.  x = 


x = 


x 3> 

a mm. 

7 r 

* = 3’ 

a max. 

a 

= ~ 2’ 

a min. 

a2 

7 > 

a max. : 

a — b 

a 2 

a min. 

a + b’ 

(a2  + x2)3 

Ans.  x = ± a,  a max. ; x = 0,  a min. 
(x  + a)  (x  + b) 


Ans.  x = — Vab,  a max. ; x = Vab,  a min. 

Ans.  x = k > 


23.  y = sin  x sin  (a  — x). 


a max. 


410 


MISCELLANEOUS  EXAMPLES. 


24. 

1 + tan  x 

7 T 

y = 

X = -T  > 

a 

min. 

sin  x 

4 

25. 

y = 

= a sec  x + b cosec  x. 

x = tan-1  1/^, 

V a 

a 

min. 

26. 

y = 

X 

Ans.  x = e, 

a 

min. 

log  X 

Find  the  minimum  value  of  each  of  the  three  following 
expressions : 

27.  a?  cosec2  9 + b2  sec2  6.  Ans.  (o  + b)2. 


28.  a tan  6 + b cot  9. 


2 Vafe. 


29.  ae™  + be-™.  2 \fab. 

V3 

30.  Find  the  max.  value  of  f sin3  9 cos  9.  — — . 

d 8 

• 

31.  The  lower  corner  of  a leaf,  width  a,  is  folded  over 

so  as  to  reach  the  inner  edge  of  the  page : find  the  width 
of  the  part  folded  over,  (1)  when  the  length  of  the  crease 
is  a minimum,  and  (2)  when  the  area  of  the  triangle  folded 
over  is  a minimum.  Ans.  (1)  fa ; (2)  fa. 


32.  A vertical  flagstaff  is  of  two  pieces,  the  upper  is  9 
feet  and  the  lower  is  16  feet  long : find  the  horizontal  dis- 
tance from  the  foot  of  the  staff  to  where  the  visual  angle 
subtended  by  the  upper  piece  is  a maximum. 

Ans.  20  feet. 


33.  A perpendicular  is  let  fall  from  the  centre  on  a tan- 
gent to  an  ellipse : find  (1)  the  length  of  this  perpendicular 
so  that  the  intercept  between  the  point  of  contact  and  the 
foot  of  the  perpendicular  is  a maximum,  and  (2)  the  inter- 
cept. Ans.  (1)  perpendicular  = Va6; 

(2)  intercept  = a — b. 


34.  The  portion  of  the  tangent  to  an  ellipse  intercepted 
between  the  axes  is  a minimum : find  its  length. 

Ans.  a + b. 


MISCELLANEOUS  EXAMPLES. 


411 


35.  Two  points,  A and  B,  5 feet  apart,  are  on  the  same 
side  of  a straight  line  CD,  and  distant  from  it  4 feet  and 
one  foot  respectively.  A straight  line  is  drawn  from  each 
of  the  points  A and  B to  a point  P on  the  line  CD.  Deter- 
mine the  length  of  the  minimum  line  APB.  Ans ■ 6.4  feet. 

36.  Being  given  the  slant  height  of  a right  circular  cone, 
find  the  cosine  of  the  semi-vertical  angle  when  the  volume 

of  the  cone  is  a maximum.  Ans.  — — • 

V3 

37.  Find  the  length  of  the  shortest  normal  chord  in  the 

parabola  y 2 = 2 px.  Ans.  3 p V3. 

38.  Determine  the  cone  of  minimum  volume  that  can  be 
described  about  a given  sphere. 

Ans.  The  sine  of  the  semi-vertical  angle  = -J. 

39.  Determine  the  cone  inscribed  in  a given  sphere  sc 
that  its  whole  surface  shall  be  a maximum. 

Ans.  Height  = (23  — Vl7). 


40.  The  sides  of  a rectangle  are  a and  h.  Find  the  maxi- 
mum rectangle  that  can  be  drawn  so  as  to  have  its  sides 
passing  through  the  corners  of  the  given  rectangle. 

Ans.  A square,  each  side  = a 

V2 


41.  If  a rectangular  piece  of  pasteboard,  the  sides  of 
which  are  a and  b,  have  a square  cut  out  at  each  corner, 
find  the  side  of  the  square  that  the  remainder  may  form  a 
box  of  maximum  volume. 


Ans. 


The  side  = 


a + b — Va2  — ah  + b2 
~6~ 


42.  A Norman  window  consists  of  a rectangle  sur- 
mounted by  a semicircle.  Given  the  perimeter,  required 


412 


MISCELLANEOUS  EXAMPLES. 


the  height  and  breadth  of  the  window  when  the  quantity 
of  light  admitted  is  a maximum. 

Ans.  The  radius  of  the  semicircle  must  equal  the  height 
of  the  rectangle. 

43.  A straight  line  drawn  from  the  extremity  of  the 
minor  axis  of  an  ellipse  cuts  the  major  axis  at  Q and  the 
curve  at  P ; from  P the  ordinate  PH  is  drawn  to  the  major 
axis ; find  when  the  area  PQH  is  a maximum. 

Ans.  PN  = |(Vl7-l). 


CHAPTER  IX. 


Find  the  equation  of  the  tangent  to  each  of  the  six 
following  curves  : 

(1.  2). 

(-2,  8). 

(2,  2). 

(2,  2). 


1.  xy  = 1 + x3,  at 

2.  y (1  + x)  = x3,  at 

3.  x2  + y2  — : 


Ans. 


:3,  at 

4.  x3  + y3  = x 4,  at 

5.  3 (x2  — x)  + 2 (y  + l)2  = 0, 

at  (1,  - 1). 


x — y + 1 = 0. 
4x  + y = 0. 
2x  — y = 2. 
5x  — 3y  = 4. 


x — 1. 

5x  _2 y = 3 
x'  y' 


6.  x5  = a2y2,  at  (x',  y'). 

Find  the  equations  of  the  tangent  and  normal  to  each  of 
the  three  following  curves  : 


7.  ay 2 = x3,  at  (a,  a). 

Ans.  3x  — 2y  = a ; 2x  + 3 y = 5a. 

8.  y2  {2a  x)  — x3,  at  (a,  a). 

Ans.  y = 2x  — a ; 2y  + x = 3a. 

9.  y (x2  + 4a2)  = 8a3,  at  (2a,  a). 

Ans.  x + 2y  = 4a ; y + 3a  = 2x. 
Find  the  lengths  of  the  subtangent,  subnormal,  tangent, 
and  normal  in  each  of  the  six  following  curves  : 


MISCELLANEOUS  EXAMPLES. 


413 


O O 

10.  ay 2 = x3,  at  (a,  a).  Ans.^a,  ^ a , ^ Vl3,  |Vl3. 

11.  4:xy=  4 + a;3,  at  (2,  1^). 

12.  y(l+x)  = x3,  at  (1,  £). 


21i  2i  is 

S'' 


5 V41  V41 


8 10 


8 


13.  y2  = 3x  + l,  at  (5,  4). 

14.  y2=  4 ax,  at  ( x , y). 

15.  yn=an~1x,  at  (x,y). 


10f,  H,|V73,^V73. 

2x,  2a. 

,2 


nx, 


r , 

nx 


16.  Find  the  length  of  the  subtangent  to  the  curve 


xmyn  = am  + n,  at  (x,  y). 


Ans.  — 


nx 

m 


17.  Find  the  length  of  the  subtangent  to  the  curve 

6 6 

r = a (1  — cos  6).  Ans.  2 a tan  - sin2 

18.  In  the  curve  x2y2  = a3  (x  + y),  find  the  angle  which 
the  tangent  at  the  origin  makes  with  the  axis  of  x. 

Ans.  135°. 

19.  In  the  curve  x2  (x  + y)  = a2  (x  — y),  find  the  equa- 
tion of  the  tangent  at  the  origin.  Ans.  y = x. 

20.  In  the  parabola  x-  + ?/-  = a%,  find  (1)  the  equation 
of  the  tangent  at  ( x y'),  and  (2)  the  length  of  the  tangent 
intercepted  between  the  co-ordinate  axes. 

Ans.  (l)-^r  + -7i  = aK  (2)  a- 

x 2 y 2 

21.  In  the  hypocycloid  yi  = find  (1)  the  equa- 
tion of  the  tangent  at  {x',  y'),  and  (2)  the  length  of  the 
tangent  intercepted  between  the  co-ordinate  axes. 

y a 


x 


414 


MISCELLANEOUS  EXAMPLES. 


22.  In  Ex.  21,  find  (1)  the  length  of  the  perpendicular 
from  the  origin  to  the  tangent  at  the  point  {x  , y'),  and  (2) 
the  area  of  the  triangle  formed  by  the  axes  and  the  tangent. 

Ans.  (1)  V ax'y',  (2)  ^ V ax' y‘ '. 


23.  In  the  curve  x4  + y4  = a4,  find  the  length  of  the 
perpendicular  from  the  origin  to  the  tangent  at  (x,  y). 

Ans.  ° 


Vx6  + ye 

24.  In  the  curve  xn  + yn  = a",  find  (1)  the  length  of  the 
perpendicular  from  the  origin  to  the  tangent  at  any  point, 
and  (2)  the  length  of  the  tangent  intercepted  between  the 
axes.  , a"  ft2" 


Ans.  (1) 


Vx2"-2  + y2n~2 


,(2)- 


y‘"  ‘ px"  1yn  1 
25.  Find  at  what  point  the  subtangent  to  the  curve 
xy2  = a2  (a  — x),  is  a maximum.  Ans. 


’•  (1’ Q) 


Find  the  asymptotes  in  the  following  curves : 


26. 

y 1 

{x2 

+ 

05 

oo 

li 

Ans.  y 

= 

X. 

27. 

y3 

6x2  — x3. 

X 

+ 

y = 

2. 

28. 

X3 

■ 

y3  = 3ftx2. 

X 

+ 

y = 

a. 

29. 

y3 

= 

6x2  + x3. 

y 

X + 

2. 

30. 

(x 

— 

2ft)  y2  = x3  — ft3. 

X 

- 

2a ; 

X 

-f-  o = 

31. 

y3 

= 

x3  + 3 ax2. 

y 

x + 

a. 

32. 

xy3  + 3 ay3  = x4. 

y 

x — 

a. 

33. 

X3 

+ 

y3  + 3ft2x  = 0. 

X 

■ 

y = 

0. 

34. 

X3 

— 

27  y3  = 2x2. 

3x  - 

= 

2. 

35. 

X4 

= 

xy3  + 3 y3. 

X 

— 

y = 

i; 

x + 

36. 

x2y  = 

= x3  + x + y. 

y 

= 

x ; x = 

= ±1. 

37. 

X3 

+ 

y3  = a3. 

X 

+ 

y = 

0. 

38. 

X3 

+ 

y3  = 3 axy. 

X 

+ 

y + 

a 

= 0. 

39. 

y2: 

c - 

- ay2  = x3  + ax2 

+ b3. 

Ans.  x = a ; y 

= x + a ; 

y 

1 + X 

+ 

a = 1 

MISCELLANEOUS  EXAMPLES. 


415 


40.  y (x2  — 3 bx  + 2b2)  = x3  — 3ax2  + a3. 

Ans.  x = b ; x = 2 b ; y + 3a  = x + 3b. 

41.  x4  — yi  = a2xy  + 52y2.  Ans.  x + 2/  = 0;  x — y = 0. 

42.  y 2 (a?/  + &x)  = a2y2  + 62x2.  y = — ^ x + 2a. 

43.  4x3  = (a  + 3x)  (x2  + y2).  y = ± (^-  - 

44.  (x  + a)  y2  = (y  + 6)  x2. 

Ans.  x + a = 0 ; y + 6 = 0 ; y + a = x + 5. 

45.  (y  — 2x)  ( y 2 — x2)  — a {y  — x)2  + 4a2  (x  + y)  = a3. 

Ans.  y = x-,y  + x = ~)  y-2x=\- 


CHAPTER  X. 

Examine  the  twenty  following  equations  for  points  of 
inflexion : 


1.  a2y  = g-  — ax2  + 2a3. 


2.  y (x2  A 3a2)  = x3. 


4.  xy  = 1 + x3. 

5.  (x  A a)2  y — a2x. 

6.  x3  — axy  = a3. 


7.  (a2  + x2)  y = a2x. 

8.  x3  + y3  = a£2. 

9.  x3  + y3  = a3. 

10.  x = y3  + 3y2. 

11.  y2  = x2  (2x  - 1). 

12.  y (a4  — 54)  = x(x  — a)4  — x64. 


Ans.  (a,  fa). 

(3a, | a), (0,0),  ( — 3a  — fa). 


(-1,  0). 

(2a,  fa). 

(a,  0). 

(0,  0),  ^±a  V3,  ±-  V3^. 

(a,  0). 

(a,  0),  (0,  a). 

(2,  - 1),  (4,  - 2),  (0,  0V 
(f  j f ~^3),  (f  , — f V3). 
x = fa. 


416 


MISCELLANEOUS  EXAMPLES. 


13.  a4y2  = a2x 4 — x6.  Ans. 

14.  y (a2  + x2)  = x3. 

15.  ay  (x  — a)  = x2  (a  + x). 

16.  log  y = xi 

17.  y V2 a — x = a Vx. 

18.  ax2  — x2y  — a2y  = 0. 

^9.  (?/  — 2\/a2x)2  = 4ax. 

20.  x3  — 3bx 2 + a2y  = 0. 


x = ±|V27-3V33. 
6 

x = 0,  x = ± a V3. 
x = — a (y/2  — 1). 
x = 8. 


a 


Examine  tlie  nine  following  equations  for  multiple  points: 

21.  a3y2  — 2 abx2y  = x5. 

Ans.  The  origin  is  a double  point  of  osculation. 

22.  y2  = 2x2  + x3. 

Ans.  The  origin  is  a double  point,  the  tangents  at  that 
point  make  the  angles  tan-1  ± V2. 

23.  (x2  + y2)2  = a 2 (x2  — y2). 

Ans.  The  origin  is  a double  point,  the  tangents  at  that 
point  bisect  the  angles  between  the  axes. 

24.  x4  4-  2 ax2y  — ay 3 = 0. 

Ans.  The  origin  is  a triple  point,  the  tangents  making 
the  angles  0 and  tan-1  ± \[2. 


25.  x2y2  = (y  + l)2  (4  - y2). 

Ans.  A double  point  at  (0,-1)  where  p = ± — • 

26.  a4?/2  = a2x4  — x6. 

Ans.  The  origin  is  a double  point  of  osculation. 


27.  y2  ( a 2 + x2)  = x2  ( a 2 — x2). 

Ans.  The  origin  is  a double  point,  the  tangents  bisect- 
ing the  angles  between  the  axes. 


MISCELLANEOUS  EXAMPLES. 


417 


28.  a2y 2 = a2x2  — x4. 

Ans.  The  origin  is  a double  point,  p = ± 1. 

29.  y2  = re  log  (1  + x). 

Ans.  The  origin  is  a double  point,  p = ± 1. 

Examine  the  eleven  following  equations  for  cusps  and 
conjugate  points. 

30-  (y  — x)2  — x3. 

Ans.  The  origin  is  a cusp  of  the  first  kind,  p = ± 1. 

31.  y2  (a  — x)  = x3. 

Ans.  The  origin  is  a cusp  of  the  first  kind. 

32.  ( x — y)2  = (x  — l)5. 

Ans.  The  point  (1,  1)  is  a cusp. 

33.  x3  + 2x2  + 2 xy  — y2  + 5x  — 2y  = 0. 

Ans.  The  point  (-1,-2)  is  a cusp  of  the  first  kind. 

34.  ( bx  — cy)2  = (x  — a)5. 

Ans.  The  point  ^ a , — j is  a cusp. 

35.  y2  = 2x2y  + x3y  + x3. 

Ans.  The  origin  is  a cusp  of  the  first  kind. 

36.  x 4 — 2 ax2y  — axy 2 + a2y 2 = 0. 

Ans.  The  origin  is  a cusp  of  the  second  kind. 

37.  y2  = 2x2y  + x4y  — 2x4. 

Ans.  The  origin  is  a conjugate  point. 

38.  ay 2 = x (a  + x)2. 

Ans.  (—  a,  0)  is  a conjugate  point. 

39.  x4  — ax2y  + axy 2 + a2y2  = 0. 

Ans.  The  origin  is  a conjugate  point. 

40.  Trace  the  curve  y2  (x  — 1)  = x2. 

Ans.  It  has  an  asymptote  x = 1,  a min.  ordinate  at 

(2,  2)  . and  a point  of  inflexion  at  ^4,  . 


418 


MISCELLANEOUS  EXAMPLES. 


41.  Trace  the  curve  y2  = x2  — x 4. 

Ans.  The  origin  is  a double  point,  where  p = ±1. 

42.  Trace  the  curve  y2  = x2  A x3. 

Ans.  The  origin  is  a double  point,  whe  e p = ± 1 ; there 
is  a max.  ordinate  at  (—  §,  f V3),  and  a min.  ordinate  at 

(-!,  -|V3). 

43.  Trace  the  curve  x2y2  = (x  + 2)2  (1  + x2). 

Ans.  It  cuts  the  axis  of  x at  tan-1  ± ^ V5;  it  has  a 
min.  ordinate  at  x = V2,  three  asymptotes,  y = x + 2, 
y = — x — 2,  x = 0,  and  a point  of  inflexion  at  x = — 6.1. 


CHAPTER 

XI. 

In  each  of  the  eleven  following  curves  find  the  radius  of 

curvature  p at  any  point  of  the  curve 

1. 

CL  X X 

y = (ea+  e “).  Ans. 

1 

II 

Q. 

2. 

y2  (2 a — x)  = x3. 

a Vx  (8a  — 3x)s 
p ~ 3 (2a  - x)2 

3. 

y2  = 4 ax. 

O (a  + 
p w V^ 

4. 

X 

(a2  + v2)f 

y + ae  a = 0. 

P — 

ay 

5. 

CO 

H 

II 

N 

e 

(4a  + 9xRx! 
p 6a 

6. 

x2  y2  . 

a2  ~ b2=  ‘ 

(62x2  - a2)i 

p afr 

7. 

Vx  + \fy  = 2 Va. 

(x  + y)i 

P Va 

8. 

e x = sin  y. 

/>  = e-x. 

MISCELLANEOUS  EXAMPLES. 


419 


9. 

Ho 

+ 

co|to 

II 

a 

■OjtO 

p = 3 (axy)K 

10. 

(a4  + 9x4)l 

a 

<ci' 

II 

H 

6a4x 

11. 

a"  xy  = x“. 

(x2  + n2y2)% 
p n (n  — 1)  xy 

12. 

y = x4  — lx3  — 18x2, 

at  (0,  0). 

P = - 36. 

13. 

y = x3  + 5x2  + 6x, 

at  (0,  0). 

P = 22.506. 

14.  y2=  at  both  points  where  y = 0. 

Ans.  (1)  p = fa ; (2)  p = fa. 
In  each  of  the  seven  following  curves,  find  the  radius  of 
curvature  at  any  point  of  the  curve. 


15. 

r = a (1  — cos 

0). 

4 4.0 

Ans.  p = ^ a sm  ^ 

16. 

r — a (2  cos  0 - 

-1). 

a (5  — 4 cos  0)1 
^ 9 — 6 cos  0 

IT. 

r = a sec2  - • 

O 3 0 

p = 2a  sec3  ^ • 

18. 

r2  cos  20  — a2. 

x3 

P a2 

19. 

r = aene. 

p = r Vl  + n2. 

20. 

1 

II 

ao 

x (4a4  + x4)l 
p 2 a2  (4a4  — x4) 

21. 

r = a sm3  - . 

3 . , 0 

p = - a sm-  g . 

22. 

a4?/2  = a2x4  — x6,  at  (0,  0)  and  (a,  0). 

23.  Find  the  order  of  contact  of  the  two  curves 
y = x3,  and  y = 3x2  — 3x  + 1. 

Ans.  Contact  is  of  the  second  order. 


420 


MISCELLANEOUS  EXAMPLES. 


24.  Of  what  order  of  contact  are  the  two  curves 
x4  x2  1 , x3  x 

^ = 12  + 2 + 12  ’ and2/=3'+3' 

Ans.  Contact  is  of  the  third  order. 
In  each  of  the  three  following  curves  find  the  coordi- 
nates of  the  centre  of  curvature  of  the  curve  at  any  point. 


25.  y3  = a2x. 

(£  x\ 
ea  + e a) 


Ans  m-ai  + 15yi  n a<y  ~ 
Ans.  to-  ^ , n — 2ai 


to  = x — ^ Vy2  — a2,  n = 2 y. 


27.  x?  + y?  = a3-  to  = x + 3x?y?,  n =y  + 3x?y?- 

Therefore  the  equation  of  the  evolute  of  this  curve  is 
(to  + n)?  + (to  — n)%  = 2a?. 

In  each  of  the  nine  following  lines  and  curves  find  the 
envelope : 

28.  y = 2 mx  A to4,  where  to  is  the  variable  parameter. 

Ans.  13  y3  + 27x4  = 0. 

29.  y2  = a(x  — a),  where  a is  the  variable  parameter. 

Ans.  Ay2  = x2. 

30.  - + \ = 1,  and  an  + bn  = where  a and  b are  the 
a b 

n n n 

variable  parameters.  Ans.  xnJrl  + yn+l  = &n  + x. 

31.  ^ + ~=  1,  and  a + b — k,  a and  b being  the  vari- 
a2  b- 

able  parameters.  Ans.  x?  + y?  — 0. 


32.  - +-  = 1,  and  a?  + 5?  = 0,  a and  b being  the  vari- 
a b 


able  parameters. 


Ans.  x?  + y?  = 0. 


MISCELLANEOUS  EXAMPLES. 


421 


a and  b are  the  variable  parameters. 


^9  + S"*- 

o? 

34.  y2  — - (x  — a),  where  a is  the  variable  parameter. 

' 4 

Ans.  y2  = 

y lie 

QX2 

35.  y = x tan  a — sec2  a,  where  a is  the  variable 

y 2v2 

2v2( 


parameter.  Ans.  x2  — — h/  — J- 

36.  xy  cos  30  sin  30=  a (cos  20)1,  where  0 is  the  variable 
parameter.  Ans.  (x2  + y2)2  = a2(x2  — y2). 


422 


MISCELLANEOUS  EXAMPLES. 


INTEGRAL  CALCULUS. 


CHAPTER  I. 

Show  by  integration  that : 

1.  f dxr.  = 4 (Va+  Vi)*. 

2.  f x~^  (a^  — X*)*  dx  = — 2 (a*  — *1)1. 

3.  j* 4:  (x3  + 1)1  x2dx  = (x3  + l)n 

4.  ^ (er  + e~x)2dx  = 4 (e2*  — e-2x)  + 2x. 

5.  ^ (3e2a:  — 1)1  e21  dx  = \ ( 3e2x  — 1)1. 

6.  (x2  — a2)2  dx  = — fa2*3  + a4x. 

7.  J* (x2  - 2)1  xdx  = 4 (x2  - 2)1 

8.  T-7a^=  = _v^t. 

J Vo2  — X2 

9-  ftt-i**  «*-*>■ 

C{  1 — x2)2  dx  . , x4 

10.  / 1 = log  x — x2  + 

11.  J* (Va  — Vx)3dx  = a-rx  — 2a*l  + §a*x2  — £x*. 

12.  ^ (a2— x2)3Vxdx=2*l  ^a4x2+ jja2x4— jgX6 

13‘  / 

A i 


dx 

V4x2  - 9 


= ^ sec- 1 fx. 


MISCELLANEOUS  EXAMPLES. 


423 


14. 

15. 

16. 


IT. 


18. 

19. 


20. 


21. 

22. 

23. 


24. 


25. 


26. 


27. 

28. 


/; 


3 dx 


. tan_1x  \fl. 


5 + 7x2  V35 

/xdx  . . x2 

= 4 tan  1 — 

x4  + 4 4 


= \ sin  1 x2. 


r dx  i _ ! o 

I = — = vers  1 ox. 

J V2x  - 3x2  V3 

/vr 

f gprr-. ! “ i 41111-1  i1- 

/'- 

J a 


xdx 


dx 

9x2  + 4 = 
sin  xdx 


= — - log  (a  + 6 cos  x). 
+ 5 cos  x o 

^ cos  (30  — 1)  d0  = J sin  (30  — 1). 

^ sec2  40  d0  = 4tan  40. 

J* sin  0 sec2  6 dd  — sec  0. 

^ sin3  0 sec3  d dd  = \ tan4  0. 

h 


dd 


sin2  0 cos2  0 


= tan  0 — cot  0„ 


^ cos30  d0  = sin  0 — £ sin3  0. 

/dx  _ x 
(1  — x2)i  Vl  — x2 

rcd  + 2£dx  = x Va2  + x2. 
J Va2  4-  x2 


424 


MISCELLANEOUS  EXAMPLES. 


29. 

30. 

31. 

32. 

33. 

34. 

35. 

36. 

37. 

38. 

39. 

40. 

41. 

42. 

43. 


/; 

r x2c 

J a - 


xdx  x 

(8x  + x2)2  4 V8x  + x2 

x2dx  x3 


(1  — x2)2  3(1  — x2)'2 

xdx  2x 


(3x  + 2x2)2  3 V3x  + 2x2 

dx  Va2  — x2 


/; 

/: 

/: 

/ 

rVx2  — 

J X4 


x2  Va2  — x2 
dx 

x2  Va2  + x2 
Va2  — x2 

Vx2  - a2 


gtx 

Va2  + x2 


a‘x 


dx  = 
dx  = 


(a2  — x2)2 
3 a2x3 

(x2  - a2)2 
3 a2x3 


(a2 

- x2)^ 

II 

1 

dx 

X 

(x2 

+ a2)* 

a2  (x2  + a2)i 

dx 

V2ax  — x2 

V2ax  — 

x2  ax 

xdx 

X 

/ 

/: 

/; 

/ 

dx 

J V4  - 3x2  V3 

/ 


(2ax  — x2)2  a V2ax  — x2 

V2ax  — x2  , (2ax  — x2)2 

dx  = — 


sin 


3 ax3 

! V3. 


Va  (62  — x2)  Va 
f Vx  I 

J V5^ 


= — — sin-1  - ■ 


Vx  dx  1 . 2 3 

. ... ..  = o sin-1  — X2. 

V5  — 4x3  ^ V5 


MISCELLANEOUS  EXAMPLES. 


425 


44 

45 

46 

47 

48 

49. 

50. 
51 


f 

/: 

h 

f. 

f 


dx 

V5x4  — 

3x2 

dx 

x i (1  + 

s*)  “ 

dx 

sin  x cos5  x 

dx 

x2  + 2x 

+ 5 

dx 

V2  + x 

— X2 

dx 

= - tan~*x^. 


= log  tan  x + tan2  x + 


tan4  x 


1 

2 

= sin- 


x + 1 
1 — 

lx  - 1 


J 5x2  — 2x  + 1 

r dx 

J x2  — 

s 


“2tan" 

= - tan- 


3 

5x  — 1 


2 

x — 3 


x*  — 6x  + 11  V2 
dx 


Vl  + 3x  — a:2 


sin" 


V2 
2x  - 3 
Vl3 


CHAPTER  II. 

Integrate  the  following  expressions  : 

f(5x3  + 1)  dx  5 ^ , 1K-  , i — (x  ~ 2)41 

' J x2  - 3x  + 2 

2 P(2x2  — 3x  + 5)  d 

J x3  — 7x2  + 36 

"'x2  + 2x  — COS2  a 


Ans.  ^x2  + 15x  + log 

j 

- • Ans.  log 


2 ~ 1 (x  - l)6 

(x  — 6)tf  (x  + 2)H 


, . o , ■ , dx.  Ans.  x + ^secalog- 

X2  + 2x  + sm2  a X + 1 — COS  a 


•Ji 

P (2  + x)  dx 

J I 


(x  — 3)ff 

X + 1+  COS  a 


(x  - l)2  (x  - 2) 


Ans.  — + 4 log  — 


x — 1 


K P(3x  — 1)  dx 
J (X  - 3)3 


x — 1 
5 - 3x 


(x  - 3)2 


Ans. 


426 


MISCELLANEOUS  EXAMPLES. 


dx 

(x  - 2 f(x  + 3)2 


Ans.  — I 

2o  U 

(a;  + 2)  dx 


+ 


x + 3 


+ log 


x + 3\T^ 
x — A2/ 


(x  — l)3  (x  + 1) 


Ans.  1 


8 

Ans. 


r & 

J (x  - l)3  ( 


4 \x  — 1 (x  — 1): 
+ 5)  dx 


log 


l\i 


x + 1 


(x  — l)3  (x  + 1)  (x  + 2) 

L_+  1 + iog  ^ + 2ii. 

2(x-1)2  + 6(x-1)+  S (*  + 1)1 


9 

10 

11 

12 

13 

14 

15 


f. 


Ans.  x + log 


x - 2\t 
x + 2/  * 


x3dx 


x2  + 7x  + 12 
— 3x  4-  3)  dx 


x2  _ , . (x  + 4)04 

2 '*+  °S(x  + 3)”' 


fix2  - 

J (x  — 


(x  - 1)  (x  - 2) 

C (3x  — 1)  dx 
J x3  — x2  — 2x 

(x5  — 5x  — 3)  dx 


x + log 


- 2 


x — 1 
1 1 ,1  (x  - 2)i 

2 X + 1)4 


■J 

f 

■f 


(x2  4-  x)2 


Ans.  ^ — 2x  + 3 

2 x2  + x 


+ log  [x  (1  + x)2]. 


9 ( — x2  + 4x  + 2)  dx 


(x2  — x — 2)3 
2x 


Ans. 


(8x6  — 1)  dx 


5 2x  + 1 . x 

+ + log 


(x  - 2)2  1 2 (x  4-  l)2 


x + 1 


(2x2  - x)3 


Ans.  x — 


12x 


1 108x  - 61 


2x2 


4 (2x  - 1) 


j + log  X24  (2x  —1) 


MISCELLANEOUS  EXAMPLES. 


427 


16-  Jr 


dx 
+ x3 

1 . (x  + l)2  1 _x 

Ans.  x log  4 — 777  tan  1 


6 x2  — x + 1 \/3 


2x  - 1 
V3 


a 1 + x + x2  1 _x  /2x  + 1\ 

Ans.  77  log  z. ^ — ; — 7 4 — 7=  tan  1 — 

6 1 — 2x  + x2  V3  V V3  / 


i q A*3  ~ !)  dx 

8 J x3  + 3x 


19-/i 


Ans.  x + log 
dx 


x2  + 3\* 


(x2  + 1)  ( X 2 4-  x) 


Tins,  t log 


4 ° (x2  + 1)  (x  + l)2  2 


V3  tan-1  — • 
V3 


— — tan  1 x. 


20. 

Jl3  + 1 


Ans.  — | log  (x  + 1)  + ^ log  (x2  — x 4-  1) 


21/(^ 


dx 


(x  — 1)  (x2  + 2 


1 x — 1 
Ans.  o log 


1 , _i  x 

- tan  1 — — . 


3 Vx2  + 2 3 V2  V2 


22./ 


x2dx 


(x  + 1)  (x  — 1)  (x2  + 2) 


1 X 1-  V2  T 

Ans.  tt  log  — + -o-tan-1  — . 
b x 4 1 3 y/2 


428 


MISCELLANEOUS  EXAMPLES. 


23 

24 

25 


I 


’(x3  + x2  + 2)  dx 
(x2  + 2)2 

Ans.  — 4=  tan  + log  (x2  + 2)\ 

x2  + 2 V2  V2 


r dx 

• J ^T' 

■/ 


log(Ir^)J_  4tan_1*- 


(2x2  — 3x  — 3)  dx 
(x  — 1)  ( x 2 — 2x  + 5) 


Ans.  log  — — ^ + £ tan-1 


(x2  — 2x  + 5)* 
x 

26  r + x2 + x + 1)  ^ 

€/ 


X — 


(x  — l)2  (x2  + 2) 


Ans.  — 


+ lo 


(x-l)¥+  5 ^ 


3 (x  - 1)  ° (x2  + 2)*  9 V2 


„7  T (x3  - 6)  dx 

' J x*  + 6x2  + 8 


, a:2  + 4 3 _ x 3 _ x 

Ans.  log  , ■ + - tan  1 - tan  1 — — . 

Vx2  + 2 2 2 V2  V2 

r (5x2  - 1)  dx 
J (x2  + 3)  (x2  — 2x  + 5) 

. , x2  — 2x  + 5 5 _ x — 1 2 . x 

Ans.  log — + -tan  1 — tan-1-—  ■ 

x2  + 3 2 2 V3  V3 


28 


29 


/: 


(9x  — 10)  dx 
x2  (2x2  — 2x  + 5) 


2 1 
Ans.  - + 75  log 


x2 


, 5.  _ , 2x  — 1 

xT2lug  2f-2x  + 5 + 3 tan  3 


30 


r x’dx 

• J " 


, 1,  (x4  — 1)  ,1  . _ 2x4  + 1 

Ans.  — log  — — - + — -^tan-1 — - 

12  (x8  + x4  + 1)4  4 V3  V3 


« I iM 


MISCELLANEOUS  EXAMPLES. 


429 


+ x5  + x3  + x)  dx 


(x2  + 2)2  ( x 2 + 3) 
5 


, IQ  | l0„>2  + 2)+^, 

2 (x2  + 2)  + x2  + 3 + ° (x2  + 3)9 


32.  f-8-r- 7 s’ 

t/  z8  + z7  — z4  — X3 


, 1 1 

^4ns.  h — o h 


1 , 1 — x 

+ o log 


2x2  x 4 (x  + 1)  8 1 + x2 

+ log  ^ tan  ~ 1 — • 

x 4 x 


33.  f 
J x4  + 


+ 1 

1 . x2  + x V2  + 1 1 x V2 

^4ns.  = log  — — tan-1 

4 V2  x3  — x V2  + 1 2V2  1 — x2 


r 

34.  / - 

J x' 


x2dx 


+ 1 

. 1 , x2  — x V2  + 1 

,4ns.  - — = log 


+ 


[tan-1  (x  V2  + 1) 


35.  f 
J i 


4V2  ° x2  + x V2  + 1 2 V2 

+ tan-1  (xV2  - 1)]. 

xdx 

(1  + x)  (1  + 2x)  (1  + x2) ' 


^ns.  _ _JL_  _ log  - (1  + - — log  (1  + X2) 


36-  f: 


5 1 + 2z 


x3dx 


(1  + 2x)^S  100 


+ 50tan'1;c- 


x8  + 1 


Ans.  loS  (z4  — x2  + 1)  — ^ log  (x2  + 1) 

+ — [tan-1  (2x  — V3)  — tan_1(2x±  V3)]. 


430 


MISCELLANEOUS  EXAMPLE 8. 


CHAPTER  III. 

Prove  the  following  by  integration  : 

1.  f ~j——x  = 2x5  + 3x5  + 6x5  + 6 log  (x5  — 1). 
J x*  — xi 

x}-dx 


■fi 

■/; 

/ 


x%  + 1 
dx 

x 5 + 

(x5  + 1)  dx 
+ x* 


= fz4  — % loS  (z4  + !)• 


= — 6x _ 5 + log 


(x*  + l)6 


= — 6x  5 + 12x  A + 2 log  x 


— log  (xA  + l)2b 

5.  f * = |x^  + 2 log  + 4 tan  “ lxK 

6.  f'2  3-x~  = ~ 2 (COt  ~ 1 

7-/vfe“"i(4a  + l)(2o“I)!- 

■J; 


9 

10 

11 

12 


(x  + 1)5  — (x  + 1)5 
= 3 [(x  + l)^  + 2 (x  + 1)5  + 2 log  }(x  + 1)5—1}]. 

r(*7^-SC*  + l)+i 


J X2  ''/x 

•/: 

•/ 

/ 


+ 1 dx  = 2 (x  + 1)- 


x Vx  + 1 
x2dx 


= log 


L 

Vx  + 1 — 1 


Vx  + 1 + 1 
6x2  + 6x  + 1 
(4x  + 1)5  12  (4x  + 1)5  ’ 

x5dx  _ 3x4  — 2x2  + 2 
V2x2  + 1 _ 30 


(2x2  + 1)5. 


30 


MISCELLANEOUS  EXAMPLES. 


431 


13. 

14. 

15. 

16. 
17. 
18 

19. 

20. 
21. 
22. 

23. 

24. 

25. 

26. 
27. 


j‘x3  (a2  — x2fsdx=  £oTj(6x4  — a2x2  — 5a4)  (a2  — x2) i 

Vr2 


/: 

/: 

/ 

/: 


dx  1 . Vx2  + a2 

= = 7T-  lQg  ' / 

a2  -"ft  Vx2  + a2  + a 


x Vx2  + ~2 
dx 


x Vx2  - x 4-  2 V2 
dx  1 


1 . Vx2- 

= -1=  log 


x + 2 + x - V2 


1 Vx2 


x V2~+  x — x2 
dx 


log 


+ 2 + x + V2 


- V2“ 


x Vx2  + 2x  — 1 

f . & 

J V4x  — 3 — x2 


V2  ° V2  + 2x  + V2  - x 
= 2 tan-1  (x  + Vx2  + 2x  — 1). 

v/s. 

V x — 3 


/ 


dx 


V2  + x - x2 
dx 


- 2 cot 


2 - 


x + 1 ‘ 


/*= 

/ 

/: 

/: 


2x  — x2 
dx 


V2  + x — x2 
dx 

V3x  — 2 — x2 
dx 

Vx2  + 3x  + 1 
dx 


r"J\ 

= 2 cot-1 V 

V3  + 1 + x 

(^)- 


V3  - 1 - 


= sin" 


= sin-1  (2x  — 3). 

= log  (2x  4-3  + 2 Vx2  + 3x  + 1) 
Vx2  - 2x  + 2 


(x  — l)2  Vx2  — 2x  + 2 


- 1 


8 4-  6x 


J x2,  V x V x 

I 
f 


(2  + 3x  - 2x2)^  25  V2  + 3x  - 2x2 

x3dx  1 2a  +6x2 
fa  + 6x2)^  Va  + 6x2 


432 


MISCELLANEOUS  EXAMPLES. 


28. 

29. 

30. 

31. 


32. 


33. 

34. 

35. 

36. 

37. 

38. 


39. 


/; 

/: 


x2dx 
(a  + bx2)% 
dx 


3 a (a  + bx2)i 

a + 2 bx2 


x2  (a  + bx2)%  cdx  ( a + bx2)% 
x3  (1  + 2x2)%  dx  = TV  (1  + 2x2)%  ( 5x 2 — 1). 
dx 


f: 


+ Vx  - 1 


log  (x  + Vx  — 1) 


2 . 

—tan 

V3 


x2  dx 
x2  Vl  + x4 


= -f=  log 
V2 


, (2  \h-l  + 1\ 

V3  / 

Vl  + X4  + x V2" 


: 


1 — x2 


Assume  x — z,  etc. 

x 


/i 

/' 

f 


— a;4 


dx 


= log 


x2  Vl  + a: 
(x2  — 1)  dx 
X Vl  + X 4 
(a:2  + 1)  dx 
x Vl  + a;4 

^1  + x2  dx 


1 . . xV2 

= — — sin-1 — . 

V2  1 + X2 

1 + X2  + Vl  + X4 


= log 


- 1 + vr t 


_ 1 ^ Vl  + a^  + x4+  xV3 
x2  Vl  + x2  + x4  V3  x2  — 1 


/; 


dx 


= sin- 


x 


(1  + X4)  V(1  +x4)l-X2  (1  + X4)‘ 

Assume  x = (1  + x4)^  sin  6,  etc. 

Vl  + x4  + x V2 


£ x2dx  _ 1 [”1^ 

J (1  — x4)  Vl  + x4  4 V2  L & 


+ tan-1 


Vl  + : 
V2 


-]■ 


1 — X2 


ra  + c 

J a - 


x4)  dx 


(1  - x4 


VT^ 


MISCELLANEOUS  EXAMPLES. 


43a 


40  fVl  + x*  dx  _ 1 

J 1 - X4 


2 V2  . 


log 


Vl  + x4  + x V2 


+ tan- 


: V2 


1 -x2 


41.  f_£: 
X2  Vx4 


>]■ 

(x4  — 1)  dx  _ Vx4  + x2  + 1 


VI  + X' 


+ X2  + 1 


X 


Assume  x2  H — 5 = z2,  etc. 
x2 


42. 


/; 


(1  — x2)  dx 


Jx 

■/; 


(1  + X2)  Vx4  + X2  + 1 
(x2  — 1)  dx 


= sin" 


X 


1 + X2 


V (x3  + ax  H-  1)  (x2  + bx  + 1) 


= 2 log 


Vx2  + ax 


+ bx  4-  1 


(1  — ax2)  dx 


(1  + ax2)  Vl  + 2 bx2  + a 


V2  (6  - a) 
when  b > a, 


45.  f 

( 


j x V2  (6  — a)  + Vl  + 2&x2  + a2x4 
& 1 + ax2  7 

1 — - sin-1  (V2  (a  ~ 6)),  when  a > 6. 
a — b)  \ 1 + ax2  / 


= sin" 


V2  (a  - 6) 
dx 

(1  + X2")  [(1  + x2n)  71  - X2]i  ^ + ^ 

Put  x = (l  + x2")2'1  sin  etc. 


46-  ff£s?’Sr~  | + JK*  («■-*)• 

47.  f -w 

x V x4  + 3X2  -f  1 


Vx4  + 3x2  + 1 


434 


MISCELLANEOUS  EXAMPLES. 


48-  farh-W1^ 


V T 

1 f(l  — X3)?  — x (1  — X3)^  + X2 


-p 

^rtan_1[^^ — — — 1 • Assume  1 — x3=xV. 

V3  L x V3  J 


49. 


/; 


dx 


= hogV— Ai±i 


(1  + x)  Vl  + 3x  + 3x2  3 z 

(1  - 2*)f  - 2 (1  - 2»)4  + 221 


-ilog 


V3 


: tan- 


'2  (l-z3)*  -z~ 
2 V3 


in  which  2 = 


1 + x 


CHAPTER  IV. 

Prove  the  following  by  integration  : 

1.  f-  ,r  -=a°? 


1 - Vi 


vr^T2 


x°  Vl  — x2  2 x 2x2 

x3dx 


'■ f: 


= — (x2  — 2a2)  Va2  + x2. 


*•/, 
6 j 


Va2  + x2  3 
x3dx  _ 2a2  + 3x2 

(a2  + x2)*  _ _ 3 (a2  + x2)*  ’ 

dx  __  x x3 

(a2  + x2)*  a4  (a2  + x2)^  3a4  (a2  + x2)- 

dx 


yja? 


’■J] 


x3  Va2  — x2 
dx 


+ — lo§ 


2a2x2  2a3  Va2  — x2  + a 


_ ^ l ^ tan- 4 — < 

(a2  + x2)2  2a2  (a2  + x2)  2a3  a 


MISCELLANEOUS  EXAMPLES. 


435 


7.  J x2  Vx2  + a?dx  = ^(2x2  + a2)Vx2  + a2 


8 

9 

10 


— g-  log  (x  + Vx2  + a2). 
(15a4  — 20a2x2  + 8x4) 


r dx 

J (a2  — x2)^  15a6  (a2  — x2)^ 

r dx  (2x2  + 1)  vr=~^ 

x4  Vl  — x2  3x3 

. f —f8 . ■ = - — (3x6  + 4x3  + 8)  VT^x3. 

J Vl  - x3  45 

11.  ^ (x2  — a2)%dx  = | (2x2  — 5a2)  Vx2  — a2 

+ fa4  log  (x  + Vx2  — a2). 

12  f f dx  = _ (2ax  _ x2)i/x  M 3 -txr 
V V2ax  - x2  V2  2 / 2 a 

■ j' x V2  ax 


13 

14 

15 

16 

17 

18 

19 

20 


\/2ax  — x2 


•/ 

. J V2 ax 


V*  _ - 3a’  + **?  ~ 2x- 

o 

, a3  x 

+ -jr-  vers  1 - • 

2 a 

dx  = V2 ax  — x2  + a vers-1  - . 

a 


x2dx 


x — a 


x — a 


\/2ax  — x2  + 77-  sin-1 

2 a 


/xda 
(2a.x  — 

I 


, «=  — V2ax  — x2  + a vers-1  — 
(2ax  - x2)*  « 


x log  x dx  = log  x — ■ 


. J* x3  log2 


xrfx  = -J-  (log2  x — £ log  X + i). 


x3  log3  xdx  = j-  (log3  x — J log2  x + | log  x — -3%). 


^ x”  log 


+ 1 Tn+  1 

xdx  = — log  X — 


n + 1 


(n  + l)2 


436 


MISCELLANEOUS  EXAMPLES. 


21 

22 

23 


' S xn  ^°g: 


:3  xdx  = 

+ 


+ 1 \ , 31og2x 

log3  x ~ 

n + 1 


n + 1 
6 log  x 


6 


(n  + l)2  (n  + l)3y 

/log  xdx  x log  x i n v 

(T^  = T^-  + l0g(1-^ 


(1  - x) 

. J* xeP  dx  = (x  — 1)  e*. 


24.  f x2eaxdx  = — (x2  - — + ?*)  • 

J a \ a ay 

25.  fXeaxdx  = -(x--)- 

J a \ a) 

26.  J" e°*  Vl  — elaxdx=  ~ [e“*  Vl  - e2^  + sin-V]. 


27 

28 

29 

30 

31 

32 

33 

34 

35 

36 


log  (x  + 2)  dx  = (x2  — 4)  log  Vx  + 2 — + x. 


■/* 

flog  (x  + 1)  dx  = 2 [lQg  (x  + 1}  _ 2]_ 

riolxdx 2 (l0g.a.  + |l0ga.  + j)- 

x?  3x2 

. riog  (i  = 2 (i+  log  a+  ^ -2  vi. 

J Vx 

■ J' x tan-1  xdx  = ^ ^ - tan-1  x — | • 

J'ta.n-1  \fxdx  = (1  +x)  tan-1  Vx  — Vx. 

. sin-1  xdx  - - x sin-1  x + Vl  — x2. 

. cos  xdx  = x sin  x + cos  x. 

. J* e*  sin  xdx  = 4 e*  (sin  x — cos  x). 

J.  x2  cos  xdx  = x2  sin  x + 2x  cos  x — 2 sin  x. 


MISCELLANEOUS  EXAMPLES. 


437 


37.  J ' x sec2  xdx  = x tan  x + log  cos  x. 

38.  J' x2  sin  xdx  = 2x  sin  x + 2 cos  x — x2  cos  x. 

39.  J V sii 


e* 


sin  2xdx  = -g-  (sin  2x  — 2 cos  2x). 


40. 


C 

/ sin  xdx  = - , --  (a  sin  x — cos  x). 

J a2  + 1 

r 12.1 

41.  / e*  sin2  xdx  = ^ eP  (1  — g sm  2x  — g cos  2x). 


42. 

3. 


53. 


e 2 cos  2 dx  = e2  l sin  ^ + cos  2 ) ' 


/ 

e®*  (sin 


ax  + cos  ax)  dx  = 


e®*  sm  ax 
a 


44.  (sin  2x  — cos  2x)  dx  = ~ e3*  (sin  2x  — 5 cos  2x). 

. (a  sin  mx  — to  cos  tox) 


45.  / sin  mxdx  = e®* 


46. 

47. 


/■ 

/■ 


e®*  cos  mxdx  = e® 


a2  + to2 

(a  cos  mx  + to  sin  tox) 


a2  + to2 


Jxsil 

/* 


TO 


1 . 


sin  tox<2x  = 1 x cos  tox sin  tox 


to 


') 


48.  J"  tan2  xdx  = x tan  x + log  cos  x — ^ • 

49.  J* cot5  xdx  = — \ cot4  x + J cot2  x + log  sin  x. 

ka  A B j tan5  Z tan3  x , , 

50.  j tan6  xax  = - — ^ ^ b tan  x — x. 

51.  J tan5  5 c?x  = tan4 1 — 2 tan2  g + log  sec4 g . 

/ si, 

u 


52.  I sin  x cos  Jxdx  = g tan3x. 


<ix 


tan3  x 


sin2  x cos4  x 


+ 2 tan  x — cot  x. 


3 


438 


MISCELLANEOUS  EXAMPLES. 


. J cosec*  2 xdx  = - . 

. j tan-1  xrfx  = x tan_1x  — ( log  (1  + x-). 

/grp  gp  1 ^ 

x cos ~lxdx  = -^cos-1  x — £ (1  — x2)"  + -j-sin-1£. 

/3  -« 

X2COS_1xdx  = ^-COS'^X  — Q Vl  — X2  (x2  + 2). 

/£* 

ex  cos3  xdx  = (cos3  x + 3 cos2  x sin  x 

+ 3 cos  x + 3 sin  x). 

/€  x 

e~x  cos3  xdx  = -jq  (6  sin  x — 6 cos  x 

+ 3 sin  x cos2  x — cos3  x). 

. J ex  sin4  xdx  = ~ (5  sin4  x — 20  sin3  x cos  x 

+ 12  sin2  x — 24  sin  x cos  x + 24), 

= 3 (tan  x — cot  x) 

+ % (tan3x—  cot3  x). 


54 

55 

56 


58 


59 


60 


61. 

r dx 
J sin4  x cos' 

62. 

[‘sin2  xdx 
J cos3  x 

63. 

j* sin2  x cos4 

/cos  X 

V 8 

64. 

/Vos4  xdx 
/ sin  x 

65. 

/* cos4  xdx 
) sin3  x 

sin  x 1 ^ 1 — sin  x 

2 cos2  x + 4 °°  1 + sin  x 


+ 


cos5  x\  sin  x 
3 


+ -• 

2 + 16 


12  3 

— (-  cos  X + log  tan  ^ • 

3 \ 

jos3  x — -r  cos  x cosec2  x 


“2log  tanI‘ 


MISCELL  AN  EO  US  EXAMPLES. 


439 


66. 


/ 


cos8  xdx  - 


sin  x 


COS3  X + ^ COS3  £ 


5 . 5 

+ sm  x cos  x 4-  jg  x. 


67.  cos8  xdx  = 


sm  x / 7 ,7  35 

cos7  a;  + g cos5  ^ + 24  cos  35 


35 


+ "jg  cos  x ) + 


35 

128  X‘ 


68 

69 

70 

71 


/dx  1 , , / tt\  x 

^-jlogtan^- jj-j. 

/tan  (a;  + a)  dx  , , , , , . 

- — = x — tan  a log  (cot  a;  — tan  a) 

tan  a; 


, f 

‘ J 4 - 5s 


5 sin  a;  3 


log 


/: 


tan  ^ — 2 


2 tan  h — 1 
A 

X 


72 

73 


& 1 „ ta”2+2 

3 + 5 cos  a:  4 °&  x 0 

tan  - — 2 

■ /5-fcosx~2taJ~‘(2tanl)' 


■/ 
74  / 


3 cos  a; 

da;  1 , ,5  tan  a;  + 4 

= - tan  - 1 . 


5 + 4 sin  2a;  3 


da; 


= -^tan-1  (3  tan  x). 


5 — 4 cos  2a;  3 

75.  J* sinmx  cos”x  dx 

sin”  + xa;  cos"  ~*x  ^ n 


If  si 

nj 


sinma;  cos n~2xdx. 


m + n m + 

Sug.  Put  u = cos"-1  a;,  and  dv  = sinmx  cos  xdx. 

sin 


du  = — (n  — 1)  cos"-2 x sin  xdx,  and  v = 


+ 1 , 


m + 1 


440 


MISCELLANEOUS  EXAMPLES. 


Then  the  formula  for  integration  by  parts  (Art.  147)  gives 


sinmx  cos nxdx 


sin”  + 1 x cos” 
m + 1 


n 


m + 

n — 1 


m + 
n — 1 


m + 


i/- 

\f 


£/- 


sinm  + 2 x cos’* _ 2 xdx  = 


sin”* + 1 x cos” 


sinmx  (cos”_2x  — cos"x)  dx  = 


m + 1 


sin”*+1x  cos”  1x 


m + 1 


sin"*x  cos n~2xdx  — — \ f sin"*  x cos’*  xdx. 

m + 1 J 


Transposing  the  last  term  to  the  first  member,  and  divid- 

m + n , 

■ , we  get  equation  / 5. 


ingbym  + l 

In  like  manner  the  five  following  “formulae  of  reduc- 
tion” may  be  obtained : 


76. 


fsi 


sinm  x cos”  xdx  = — 


m + n 


1 c ■ 

- I S1I 

nj 


7YI  — 

-1 — / sinm-2x  cos nxdx. 


m + n 


sinm  xdx  = — 


sin”*  ~ 1 x cos  x m — 1 


sin”*-2  xdx. 


xdx. 


77.  /si 

/si 
cos’*  xdx  = — 

j sinmx  (m  — 1)  sin”*-1x  1 m 

/dx  _ sin  x ^ n — 2 C dx 

cos"x  ( n — 1)  cos’*_1x  n — 1 J cosn_2x 


79 

80 


sin  x cosn_1x  n — 1 
n n 

cos  x m 


^ cos” 

- 2 r dx 

— 1 J sin”*  ~ 


cos"x  ( n — 1)  cos’* 

The  results  in  these  six  formulae  hold  whether  m or  n 
be  positive  or  negative,  integral  or  fractional. 


MISCELLANEOUS  EXAMPLES. 


441 


CHAPTER  V. 


. Integrate  the  following 
9 


x3dx  = g x-  — - x^ 


• 

X 

C dx  , x6  3xu 

J VI  - X5  12  88 

. C dx  _ j-7 — , sin2x  , 

4.  I ■ : = 2 Vsin  a;  1 + + 

J Vsin  a:  \ 10 


111  1 15 

— rr  — t 2 . 

44  X 120  X 


ePdx  x2  x3  x4  x5  x6 

cosx  = X + '2+T  + '6  + 10  + 20  ' 

5x16 
88  256 

sin2x  sin4x 


24 


5 

6 

7 

8 

9 

10 

11 

12 


x2dx. 


+ x2 
8 a3dx 
x2  + 4 a2 


•X 

* */*  ^2:c  — 
f3  xdx 

• J2  r 

•I 

•f 

X' 

7T 

•r- 

• L si 


sec4  xdx. 
x log  xdx. 


sin2  xdx. 
sin3  xdx. 


13 

14 


•X 
•X‘i- 


cos5  xdx. 
dx 


Ans.  21. 

V 
4 * 


5 loS  2- 

2nd2. 


4 

3 

e2  + 1 


IT 

4* 

11 

12’ 

8^ 

15' 


x + x2 


3v£ 


442 

MISCELLANEOUS 

EXAMPLES. 

15. 

fr2V 2r  dX  • 

Ans.  Sr. 

0 

< 

to 

~s 

1 

7 r 

n 

1 

16. 

1 sm3  x cos3  xdx. 

12' 

17. 

r dx 

7 T 

Jo  (a2  + x2)  ( b 2 + x 2) 

2ab  (a  + b ) 

18. 

p dx 

, 1 + V2 

log — 

Jn  COS  x 

V3 

19. 

J*  x cos  xdx. 

a sin  a + COS  a — 

20. 

^ x sin  xdx. 

sin  a — a COS  a. 

21. 

r dx 

7 r — a 

Jo  X2  — 2x  cos  a + 1 

sin  a 

1 

22. 

/ e~°*dx. 

J 0 

a 

23. 

7 r 

sin  xdx 

£ log  2. 

Jo  COS  x 

24. 

J'  sec  xdx. 

log  (1+  V2). 

on 

C2  dx 

1 

0 (4  + 3x2)f  * 8 


26. 

J'  Va2  — x2dx. 

7r  a2 

X' 

27. 

X”Ters"‘  ($*• 

ira. 

28. 

p dx 
Jo  Va  — x 

2 Va. 

29. 

p dx 
Jo  "s/ax  — x? 

7T. 

30. 

p dx 
j 1 X Vx2  — 1 

7T 

3 

MISCELLANEOUS  EXAMPLES. 


448 


31. 

32. 

33. 

34. 

35. 

36. 

37. 

38. 

39. 

40. 

41. 

42. 

43. 

44. 

45. 


sin  1 xdx. 


£si 

r 

x 
x 

r%-  si 

Jo  n 


dx 

(1  + x)  Vl  + 2x  — x2 
'da2  — x2  cos-1  - dx. 

f o ^ 

77 

¥ 

tan5  xdx. 
sin  xdx 


+ cos2  x 

X2a  

V2ax  — x3dx. 

f 

i: 


x vers-1-  dx. 
a 


x2  vers-1  - dx. 
a 


sin2  x cos3  xdx. 


vers- 


V2 ax  — x2  vers  1 - dx. 

a 


X 
X 
X 

X2a  

x V2ax  - 
Pbr 

J tan'  xdx. 
x^dx 

Vo  Vo  — X 

X 


dx. 


x2  vers  1 - dx. 
a 


Jo  Vo  — x 
>2“  xdx 


V2ox  — x2 


7 r 

4V2' 

■n^o2  ; a2 
"16~  + 1 ' 


i (log  2 


7T  J t 

— f-  tan-1 


*)■ 

i 

W 


-ir a* 
~2 


ira2. 


U*a3. 

Tz  ■ 

(tt  — 4)  a. 

A2 

4 

4a3  t ra3 

T~  + ~T' 

A - i log  2. 


tV™ 


3 


7 ra 


444 


MISCELLANEOUS  EXAMPLES. 


46. 

47. 

48. 

49. 

50. 

51. 

52. 

53. 

54. 

55. 


56. 


57. 


(a2  — e2x2)  dx 

Va2  — x2 


Ans.  — (1  - 


r 

Jr'n  x sin  xdx 
o 1 + cos2  x ' 

J (2ax  — x2)%  vers”1  ^dc.  ^^a4. 

X 
X 

7T 

^ fa  cos^  20  cos  0rf0. 

^ V3 

. aV* 


e cos  mxdx. 

D 

e-®*  sin  mxdx. 


7 r2 

4" 


a 

a 2 + to2 
m 

a2  + to2 

7ra 

4V2’ 


a2  — x2dx. 


^f7T03. 


X 

X 

X 


dr 


2 + cos  x 


dr 


4 + 5 sin  x 


dr 


5 + 4 sin  x 

2 cos2  0 sin  0 <?0 
+ e2  cos2  0 


+ 2 CO 

Jo  T 


3 V3 
i log  2. 

f tan-1  (J). 

1 tan-1  e 
e2  e3 


MISCELLANEOUS  EXAMPLES. 


445 


CHAPTER  VI. 

LENGTHS  OF  CURVES. 

1.  Find  the  length  of  the  curve  y2  = 4 ax  between  the 

points  where  x = 0 and  x — a.  Ans.  2.29558 a. 

2.  Find  the  length  of  the  curve  ay 2 = x3  between  the 

points  where  x — 0 and  x = 5a.  Ans.  %3y5-a. 

3.  Find  the  length  of  the  curve  y = log  x between  the 

points  where  x = 1 and  x = 5.  Ans.  4.37. 

4.  Find  the  length  of  the  loop  of  the  curve 

9 ay2  = x(x  — 3a)2.  Ans.  4 a V3. 

5.  Find  the  length  of  the  hypocycloid  xA  + if  = a? 
between  the  points  where  x = — ~ and  x = a.  Ans.  2.445a. 

6.  Find  the  length  of  the  curve  ev  = 1 — x2  between 

the  points  where  x = 0 and  x = Ans.  log  3 — 

7.  Find  the  length  of  the  curve  y3  = ax2  between  the 

points  (0,  0)  and  (a,  a).  Ans.  (13?  — 8)  —■ 

8.  Find  the  length  of  the  spiral  of  Archimedes  r = ad 
between  the  points  where  0 = 0 and  6 = 6tt. 

Ans.  [67t  Vl  + 3 Ott2  + log  (67r  + Vl  + 367r2)]  — ■ 

9.  Find  the  length  of  the  curve  r = ea°  between  the 

points  where  r = 0 and  r = 2a.  Ans.  2 Va2  + 1. 

10.  Find  the  length  of  the  hyperbolic  spiral  rO  = a,  (1) 
between  the  points  where  6 = yV  an(^  $ = and.  (2)  be- 
tween the  points  where  0 = | and  0 = 

Ans.  (1)  [f  J + log  |]  a ; (2)  [V  + log  f ] a. 


446 


MISCELLANEOUS  EXAMPLES. 


11.  Find  the  length  of  the  arc  of  the  parabola  r = 

7)  _ 7j* 

3 — 7T  between  the  points  where  0 = — and  0 = n. 

1 — cos  0 2 

Ans.  ^ [^2  — log  tan  ^ ],  or  | [ V2  + log  (1  + V2)]. 

12.  Find  the  entire  length  of  the  curve  r = a sin3-^  . 

O 

Ans.  %-rra. 


CHAPTER  VII. 

PLANE  AREAS. 

1.  Find  the  area  of  a loop  of  the  curve  a2y 4 = a2xA  —x6. 

Ans.  fa2. 

2.  Find  the  area  of  both  loops  of  the  curve 

y2  = x2  (1  — x2)3.  Ans.  f . 

3.  Find  the  area  between  the  curve  y2  = x*  + x5  and 

the  axis  of  y,  at  the  left  of  the  axis  of  y.  Ans. 

4.  Find  the  area  inclosed  by  the  curve  y ( a 2 + x2)  = 
(a  — x)  and  the  axis  of  x,  between  the  points  where 

x = 0 and  x = a.  Ans.  4 log  2 — 

4: 

6.  Find  the  area  inclosed  by  the  curve  y = Ax  — x3  and 
the  axis  of  x,  between  the  points  where  x = — 2 and  x —2. 

Ans.  8. 

6.  Find  the  area  between  the  curve  (?/  — x)2  = x3  and 

the  axis  of  x.  Ans.  XV 

7.  Find  the  area  between  the  curve  ( y — x2)2  = x5  and 

the  axis  of  x.  Ans.  vV 

8.  Find  the  area  between  the  curve  a2y  = x (x2  — a2) 

a2 

V' 


and  the  axis  of  x. 


J.ns. 


MISCELLANEOUS  EXAMPLES. 


447 


9.  Find  the  area  between  the  curve  y (1  + x2)  = 1 and 
the  axis  of  x.  Ans.  it. 


10.  Find  the  area  between  the  curve  y = x(l  — x2)  and 

the  axis  of  x.  Ans. 

11.  Find  the  area  between  the  curve  y = x2  (x  — 1)  and 

the  axis  of  x.  Ans.  XV 

12.  Find  the  area  of  the  loop  of  the  curve  y2  = x*(2x  + 1). 

Ans.  x$x. 

13.  Find  the  area  of  the  loop  of  the  curve  y2  = x2(2x  + 1)- 

Ans.  XV 

14.  Find  the  area  of  the  loop  of  the  curve 

8 


ay2  — (x  — a)  (x  — 2a)2. 

15.  Find  the  area  of  the  loop  of  the  curve 


Ans.  -rva2- 
15 


a3y2  = x4  (b  + x). 


Ans. 


32  h 2 
105a  i 


X . X 

16.  Find  the  area  between  the  curve  v — sin -log  sin-, 

a a a 

the  x axis,  and  the  ordinates  whose  abscissas  are  0 and  air. 

Ans.  2a  (1  — log  2). 

17.  Find  the  area  of  the  ellipse  ax2  + 2 bxy  + cy2  = 1. 

Ans.  ir 


Vac  — b2 

18.  Find  the  area  between  the  cissoid  x3  = y2  (a  — x) 

and  its  asymptote.  Ans.  |7ra2. 

19.  Find  the  area  between  the  curves  y2  = 2 px,  and 

x2  = 2py.  Ans.  |-p2. 

20.  Find  the  area  between  the  curves  y (1  + x2)  = x,  and 

Ans.  log  4 — f . 

21.  Find  the  area  between  the  curves  y2  = Sx,  and 

y = 2x  — 8.  Ans.  36. 


x 

y = i 


448 


MISCELLANEOUS  EXAMPLES. 


22.  Find  the  area  between  the  curve  y 2 (2a  — x)  = x3 

and  its  asymptote.  Ans.  ra2. 

23.  Find  the  area  between  the  curves  x2  = 4ay,  and 

y ( x 2 + 4a2)  = 8a3.  Ans.  (2 tt  — f ) a2. 

24.  Find  the  area  between  the  curves  x = 2 vers-1  ^ — 

g^. 

V4 j/  — y2,  and  y2  Ans.  § tt. 

7 r 

25.  Find  the  area  of  the  loop  of  the  curve 


16a4y2  = b2x2{a2  — 2 ax). 


Ans. 


ab 

30' 


26.  Find  the  area  of  a loop  of  the  curve  2y2  ( a 2 + x2)  = 

(. a 2 - x2)2.  Ans.  a 2 [3  V2  log  (1  + V2)  - 2], 

27.  Find  the  area  of  a loop  of  the  curve 

y2  ( a 2 + x2)  = x2(a2  — x2).  Ans.  ~ (tt  — 2). 

28.  Find  the  area  of  a loop  of  the  curve  2y 2 (a2  + x2)  — 
4ay  (a2  — x2)  + (a2  — x2)2  = 0.  Ans.  a2ir  (4  — ^ V2). 

29.  Find  the  area  between  the  curve  r = e*6  and  the 


initial  line,  from  0 = 0 to  6 = 


Ans.  -j — (e™  — 1). 
4a 


30.  Find  the  area  of  a loop  of  the  curve  r = a sin  26. 

. 7 TO2 

Ans.  -Q- 

O 


31.  Find  the  area  between  the  curve  r = a sec2  - and 


the  initial  line,  from  6 = 0 to  6 = ^ • 


A?is.  fa2. 


32.  Find  the  area  of  a loop  of  the  curve  r cos  6 = a cos  26. 


MISCELLANEOUS  EXAMPLES. 


449 


33.  Find  the  area  of  a loop  of  the  curve 

r3  cos  6 = a?  sin  36.  Ans.  f a2  — E log  2. 

u 

Q 

34.  Find  the  area  between  the  curve  r = a sin3  ^ and 

o 

the  initial  line,  below  the  initial  line. 

Ans.  (IOtt  + 27  VS) 


35.  Find  the  area  of  the  loop  of  the  curve  r = 2 cos  6 + 3. 

Ans.  II7 r. 


36.  Find  the  area  of  the  loop  of  the  curve 
3a  cos  6 sin  6 
sin3#  + cos3# 


Ans.  | a2. 


CHAPTER  VIII. 

SURFACES.  VOLUMES. 

1.  Find  (1)  the  area  of  the  surface,  and  (2)  the  volume 
of  the  solid,  generated  by  the  revolution  of  y2  — 4=ax  about 
the  axis  of  x from  x = 0 to  x = a. 

Ans.  (1)  f ( V8  -1)  tto2  ; (2)  2^-u3. 

2.  Find  (1)  the  area  of  the  surface,  and  (2)  the  volume 
of  the  solid,  generated  by  the  revolution  of  the  curve 
xi  + y3  = a ! about  the  axis  of  x. 

Ans.  (1)  -VW;  (2) 

3.  Find  (1)  the  area  of  the  surface,  and  (2)  the  volume 
of  the  solid,  generated  by  the  revolution  of  the  curve 
x~  + (y—b)2  = a2  about  the  axis  of  x,  b>  a. 

Ans.  (1)  i-T+ab',  (2)  2n2a2b. 


450 


MISCELLANEO  US  EX  AMP  L ES. 


4.  Find  (1)  the  area  of  the  surface,  and  (2)  the  volume  of 
the  solid,  generated  by  the  revolution  of  the  curve  y = 
a / x _x\ 

2 \e°  + e a)  about  the  axis  of  y from  x = 0 to  x = a. 

HTfl  ^ 

Ans.  (1)  2to2  (1  - e-1) ; (2)  ™ (e  + 5tr*  - 4). 


5.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  the  area  between  the  curve  y2  (a2  — x2)  = a 4 and 
its  asymptote,  (1)  about  the  axis  of  y,  and  (2)  about  its 
asymptote.  .4ns.  (1)  47m3;  (2)  2?ra3  (? r — 2). 


6.  Find  the  volume  of  the  solid  generated  by  the  revo- 


lution of  the  curve  ^-Y  + 


w 

b, 


— 1 about  the  axis  of  x. 

Ans.  -ff;Trab2. 


7.  Find  (1)  the  area  of  the  surface,  and  (2)  the  volume  of 
the  solid,  generated  by  the  revolution  of  the  curve  y = e~ x 
about  the  axis  of  x from  x = 0 to  x = oo  . 

Ans.  (1)  [ V2  + log  (1  + V2)>;  (2) 


8. 


Find  the  surface  of  the  oblate  spheroid. 


Ans. 


27m2  + — log 
e 


9.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  the  area  between  the  witch  y2  (a  — x)  — a2x  and 
its  asymptote,  (1)  about  its  asymptote,  and  (2)  about  the 
axis  of  y.  Ans.  (1)  ^7r2a3 ; (2)  |7ra3. 


10.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  the  area  between  the  cissoid  y2  (a  — x)  = .r3  and 
its  asymptote,  (1)  about  its  asymptote,  and  (2)  about  the 
axis  of  y.  Ans.  (1)  J (2)  j ^ a 3. 


MISCELLANEOUS  EXAMPLES. 


451 


11.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  the  closed  portion  of  the  curve  ( y 2 — b2)2  = a3x 

256  7T b9 


round  the  axis  of  y. 


MS. 


12.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  the  curve  ( x 2 + y 2)2  = a2x2  + b2y2,  (1)  round  the 
axis  of  x,  and  (2)  round  the  axis  of  y,  a > b. 

Vo2  — b 2 + a 


TT(l 


Ans.  (1)  ~ (2a2  + 3b2) 

7t6 


7t54 

2 Va2  - b2 


log 


(2)  -g-  (3a2  + 262)  + 


ira'* 


2 Va2 


: Sill' 


Va2  - b2 


13.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  a loop  of  the  curve  r = a cos  26  about  its  axis. 

^(8V2-9, 


Ans. 


14.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  a loop  of  the  curve  r = a sin  26  about  the  initial 


line. 


Ans.  yW  ’ra3 


15.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  a loop  of  the  lemniscate  r2  = a2  cos  26  about  its 


axis. 


. ml 

.4  ns.  — 
4 


-4  log  (1  + V2)  - i 

,V2  3. 


16.  Find  the  volume  of  the  solid  generated  by  the  revo- 
lution of  a loop  of  the  lemniscate  r2  = a2  sin  26  about  the 

Tr2a3 

IT  ‘ 


initial  line. 


Ans. 


A Selection  of  Reference  and  Text  Boo\s 
From  the  Van  Tfostrand  'M.ailxematics  List 



An  Elementary  Treatise  on  Analytic  Mechanics, 

By  E.  a.  bowser.  $3.00 

Industrial  Mathematics,  By  paul  v.  farnsworth. 

$2.50 

Mathematics  for  the  Practical  Man,  By  george 
howe.  $1.50 

Handbook  of  Engineering  Mathematics,  By  Wal- 
ter e.  Wynne,  b.e.,  and  w.  spraragen,  b.e.  Second 
Edition — Third  Printing.  $2.50 

Rapid  Arithmetic,  By  t.  o’connor  sloane.  $1.50 

Elementary  Mathematics  for  Engineers,  By 

ERNEST  H.  SPRAGUE.  $2.00 

Plane  Trigonometry,  By  p.  r.  rider  and  Alfred 
davis.  $1.90 

Elementary  Treatise  on  Analytic  Geometry,  By 
E.  A.  BOWSER.  $1.75 

A Course  in  Analytic  Geometry,  By  p.  p.  boyd, 
j.  m.  davis  and  E.  l.  rees.  $2.40 

Descriptive  Geometry,  By  william  s.  hall.  $2.50 

Vector  Calculus,  By  james  b.  shaw.  $3.50 

Applied  Calculus,  By  robert  g.  thomas.  $3.00 

Elements  of  the  Differential  and  Integral  Calcu- 
lus, with  Applications,  By  w.  s.  hall.  $2.75 

A New  Manual  of  Logarithms  to  Seven  Places 
of  Decimals,  By  dr.  bruhn.  Cloth  $3.00 

Flexible  $4.00 

Logarithmic  Tables  of  Numbers  and  Trigonomet- 
rical Functions,  By  baron  von  vega.  $2.00 


Complete  Description,  or  examination  copies, 
sent  on  request. 


Duke  University  Libraries 


D00349209R 


IIIIIIIIIIIIIIIIIIIIIIHI 


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